Circles and Angular Relationships

Circle Anatomy is about learning the names and meanings of the important parts of a circle. When you understand these parts, it becomes much easier to solve problems involving circles, arcs, and angles.

A circle is the set of all points in a plane that are the same distance from one fixed point. That fixed point is called the center.

The distance from the center to any point on the circle is called the radius. Every radius in the same circle has the same length.

The most important relationship in circle anatomy is between the radius and the diameter. A diameter is a line segment that passes through the center of the circle and has endpoints on the circle. Because it is made of two radii placed end to end, the diameter is always twice the radius.

We can write this relationship as:

$$d = 2r$$

and also

$$r = \frac{d}{2}$$

Here, \(r\) stands for radius and \(d\) stands for diameter.

Now let’s look at other important parts of a circle.

• Chord: a line segment with both endpoints on the circle.
• Diameter: a special chord that passes through the center.
• Secant: a line that cuts through the circle at two points.
• Tangent: a line that touches the circle at exactly one point.
• Point of tangency: the single point where a tangent touches the circle.

It helps to compare these parts carefully. A chord is only the segment inside the circle between two points on the circle. A secant is a full line that continues in both directions and passes through the circle, crossing it at two points. A diameter is both a chord and a special segment because it goes through the center.

A tangent line is different from a secant line. A tangent only touches the circle once, while a secant crosses through it.

There is one very important fact about tangents: a tangent is perpendicular to the radius drawn to the point of tangency.

That means the angle formed is a right angle, or \(90^\circ\).

$$\text{radius} \perp \text{tangent}$$

This fact is used often in geometry problems.

Here is a quick review of the main circle parts:

• Center: the fixed middle point of the circle
• Radius: from center to circle
• Diameter: across the circle through the center
• Chord: between two points on the circle
• Secant: a line crossing the circle twice
• Tangent: a line touching the circle once

Important comparisons:

• Every diameter is a chord.
• Not every chord is a diameter.
• A secant is a line; a chord is a segment.
• A tangent touches once; a secant intersects twice.

Let’s work through some examples.

Example 1: Find the diameter from the radius

A circle has radius \(6\) cm. Find the diameter.

Use the formula:

$$d = 2r$$

Substitute \(r = 6\):

$$d = 2(6) = 12$$

Answer: The diameter is \(12\) cm.

Example 2: Find the radius from the diameter

A circle has diameter \(18\) inches. Find the radius.

Use the formula:

$$r = \frac{d}{2}$$

Substitute \(d = 18\):

$$r = \frac{18}{2} = 9$$

Answer: The radius is \(9\) inches.

Example 3: Identify the circle part

Suppose a line crosses a circle and meets it at two points. Is this line a chord, secant, or tangent?

Because it is a line that intersects the circle at two points, it is a secant.

If only the piece between the two intersection points were shown, that piece would be a chord.

Answer: The line is a secant.

Example 4: Use the tangent and radius relationship

A tangent line touches a circle at point \(T\). A radius is drawn from the center \(O\) to point \(T\). What is the measure of angle \(OT\) with the tangent line?

The radius to the point of tangency is perpendicular to the tangent line.

Perpendicular lines form a right angle.

$$\angle OT\text{ with tangent} = 90^\circ$$

Answer: The angle measures \(90^\circ\).

How to recognize each part in a diagram

1. Find the center first.
2. If a segment goes from the center to the circle, it is a radius.
3. If a segment goes across the circle through the center, it is a diameter.
4. If a segment connects two points on the circle but does not need to pass through the center, it is a chord.
5. If a full line crosses the circle at two points, it is a secant.
6. If a line touches the circle once, it is a tangent.

Common mistakes to avoid

• Do not confuse radius and diameter. The diameter is twice as long as the radius.
• Do not call every chord a diameter. A diameter must pass through the center.
• Do not confuse a secant with a chord. A secant is a line, while a chord is only a segment.
• Do not forget that a tangent touches the circle at exactly one point.
• Remember that the radius and tangent form a \(90^\circ\) angle at the point of tangency.

Practice questions

1. A circle has radius \(11\) cm. What is its diameter?
2. A circle has diameter \(24\) m. What is its radius?
3. Is a diameter also a chord?
4. A line touches a circle at exactly one point. What is it called?
5. A line crosses a circle at two points. What is it called?

Answers:

1. \(22\) cm
2. \(12\) m
3. Yes
4. Tangent
5. Secant

Summary

Circle anatomy means knowing the parts of a circle and how they relate. The radius goes from the center to the circle, and the diameter goes across the circle through the center, with relationship \(d = 2r\). A chord connects two points on the circle, a secant crosses the circle at two points, and a tangent touches the circle once. Also, a radius to the point of tangency is always perpendicular to the tangent line.

Lesson: Circumference, Arc Length, and Introduction to Radians

Circles appear everywhere: wheels, clocks, coins, and round tracks. To understand circles well, we need to know how to measure the distance around them and how to measure parts of that distance.

In this lesson, you will learn three connected ideas:

• circumference: the total distance around a circle,
• arc length: the distance along part of a circle,
• radians: another way to measure angles besides degrees.

These ideas fit together because a full turn around a circle is linked to both angle measure and distance along the circle.

1. Circumference of a Circle

The circumference is the perimeter of a circle. It is the total distance around the outside.

Two important parts of a circle are:

• radius \, the distance from the center to the circle, written as \(r\),
• diameter \, the distance across the circle through the center, written as \(d\).

The diameter is always twice the radius:

$$d = 2r$$

The formula for circumference is:

$$C = 2\pi r$$

Since \(d = 2r\), the circumference can also be written as:

$$C = \pi d$$

Here, \(\pi\) (pi) is a special number that is about \(3.14\).

Example 1: Finding circumference

A circle has radius \(6\text{ cm}\). Find its circumference.

Use the formula:

$$C = 2\pi r$$

Substitute \(r = 6\):

$$C = 2\pi(6) = 12\pi$$

So the exact circumference is:

$$12\pi\text{ cm}$$

As a decimal, this is about:

$$12\pi \approx 37.68\text{ cm}$$

2. Arcs and Arc Length

An arc is a part of the circle’s edge. If you draw two radii from the center, they cut off a piece of the circle. That curved piece is an arc.

The arc length is the distance along that curved part.

If the angle at the center is a fraction of a full circle, then the arc length is the same fraction of the full circumference.

A full circle measures \(360^\circ\). So if the central angle is \(\theta\) degrees, then:

$$\text{arc length} = \frac{\theta}{360^\circ} \cdot 2\pi r$$

This formula makes sense because:

• \(\frac{\theta}{360^\circ}\) tells what fraction of the whole circle you have,
• \(2\pi r\) is the whole circumference.

Example 2: Arc length using degrees

A circle has radius \(10\text{ cm}\), and the central angle is \(72^\circ\). Find the arc length.

Use the formula:

$$\text{arc length} = \frac{\theta}{360^\circ} \cdot 2\pi r$$

Substitute the values:

$$\text{arc length} = \frac{72}{360} \cdot 2\pi(10)$$

Simplify \(\frac{72}{360} = \frac{1}{5}\):

$$\text{arc length} = \frac{1}{5} \cdot 20\pi = 4\pi$$

So the arc length is:

$$4\pi\text{ cm}$$

As a decimal:

$$4\pi \approx 12.56\text{ cm}$$

3. Why We Need Another Angle Measure

You already know degrees. A full circle is \(360^\circ\). Degrees are useful, but in circle work there is another angle unit called the radian.

Radians connect angle measure directly to the radius and the arc length. This makes some circle formulas simpler.

4. What Is a Radian?

A radian is based on arc length.

Imagine a circle with radius \(r\). If an angle at the center cuts off an arc whose length is also \(r\), then that angle measures 1 radian.

So a radian is not chosen randomly. It comes from the circle itself.

Key idea:

$$1\text{ radian means arc length } = \text{radius}$$

5. Connecting Degrees and Radians

A full circle has circumference \(2\pi r\). Since one radian is the angle that cuts off arc length \(r\), the number of radians in a full circle is:

$$\frac{2\pi r}{r} = 2\pi$$

So:

$$360^\circ = 2\pi \text{ radians}$$

Half of that is:

$$180^\circ = \pi \text{ radians}$$

This is the most important degree-radian relationship.

From it, we get common angle conversions:

• \(180^\circ = \pi\) radians
• \(90^\circ = \frac{\pi}{2}\) radians
• \(60^\circ = \frac{\pi}{3}\) radians
• \(45^\circ = \frac{\pi}{4}\) radians
• \(30^\circ = \frac{\pi}{6}\) radians
• \(360^\circ = 2\pi\) radians

To convert degrees to radians:

$$\text{radians} = \text{degrees} \cdot \frac{\pi}{180}$$

To convert radians to degrees:

$$\text{degrees} = \text{radians} \cdot \frac{180}{\pi}$$

Example 3: Convert degrees to radians

Convert \(120^\circ\) to radians.

Use:

$$\text{radians} = \text{degrees} \cdot \frac{\pi}{180}$$

$$120 \cdot \frac{\pi}{180} = \frac{120\pi}{180}$$

Simplify:

$$\frac{120\pi}{180} = \frac{2\pi}{3}$$

So:

$$120^\circ = \frac{2\pi}{3}\text{ radians}$$

6. Arc Length Formula Using Radians

When the angle is measured in radians, the arc length formula becomes much simpler:

$$s = r\theta$$

Here:

• \(s\) is the arc length,
• \(r\) is the radius,
• \(\theta\) is the angle in radians.

This formula works because radians are built from the relationship between radius and arc length.

Be careful: this formula only works directly when \(\theta\) is in radians, not degrees.

Example 4: Arc length using radians

A circle has radius \(8\text{ cm}\). A central angle measures \(\frac{3\pi}{4}\) radians. Find the arc length.

Use:

$$s = r\theta$$

Substitute the values:

$$s = 8\left(\frac{3\pi}{4}\right)$$

Simplify:

$$s = 6\pi$$

So the arc length is:

$$6\pi\text{ cm}$$

As a decimal:

$$6\pi \approx 18.84\text{ cm}$$

7. Comparing the Two Arc Length Formulas

You can find arc length in two main ways:

• If the angle is in degrees:

$$\text{arc length} = \frac{\theta}{360} \cdot 2\pi r$$

• If the angle is in radians:

$$s = r\theta$$

The radian formula is shorter and often easier, but only if the angle is already in radians.

8. Common Mistakes to Avoid

• Using diameter instead of radius by mistake. In \(C = 2\pi r\) and \(s = r\theta\), make sure \(r\) is the radius.
• Forgetting that arc length is part of the circumference. It should be smaller than the whole circumference unless the angle is a full circle.
• Using \(s = r\theta\) with degrees. That formula only works when \(\theta\) is in radians.
• Not simplifying angle conversions. For example, \(\frac{90\pi}{180} = \frac{\pi}{2}\), not just \(\frac{90\pi}{180}\).

9. Quick Check Ideas

Ask yourself these questions when solving:

• Am I finding the whole distance around the circle or just part of it?
• Is the angle given in degrees or radians?
• Do I know the radius or the diameter?
• Does my answer make sense compared to the full circumference?

10. Lesson Summary

The circumference of a circle is the full distance around it, found with \(C = 2\pi r\) or \(C = \pi d\).

An arc length is part of that distance. If the angle is in degrees, use:

$$\text{arc length} = \frac{\theta}{360} \cdot 2\pi r$$

A radian is an angle that cuts off an arc equal in length to the radius. The key relationship is:

$$180^\circ = \pi \text{ radians}$$

When the angle is in radians, arc length is especially simple to find:

$$s = r\theta$$

Understanding how circumference, arcs, and radians connect will help you solve many circle problems more easily.

Area of Sectors and Segments

When a circle is cut into parts, those parts can have different names and different area formulas. In this lesson, you will learn how to find the area of a sector and the area of a segment.

This is useful when working with pizza slices, clock faces, pie charts, and many geometry problems involving circles.

Before we begin, remember the area of a whole circle:

$$A = \pi r^2$$

Here, \(r\) is the radius of the circle.

1. What is a sector?

A sector is a part of a circle formed by two radii and the arc between them. It looks like a slice of pizza.

If the central angle is small, the sector is a small part of the circle. If the central angle is large, the sector takes up more of the circle.

Because a sector is just a fraction of the whole circle, its area is the same fraction of the circle's total area.

Area of a sector

If the central angle is \(\theta\) degrees, then:

$$\text{Area of sector} = \frac{\theta}{360} \cdot \pi r^2$$

This works because a full circle is \(360^\circ\).

2. What is a segment?

A segment is the region between a chord and the arc it cuts off.

A segment is not the same as a sector. A sector is bounded by two radii and an arc. A segment is bounded by a chord and an arc.

To find the area of a segment, we usually subtract the area of a triangle from the area of a sector.

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$

This triangle is formed by the two radii and the chord.

3. Steps for finding sector area

1. Find the radius \(r\).

2. Find the central angle \(\theta\).

3. Use the formula $$\frac{\theta}{360}\pi r^2$$

4. Simplify the answer. Leave in terms of \(\pi\) unless told to round.

4. Steps for finding segment area

1. Find the area of the sector.

2. Find the area of the triangle inside the sector.

3. Subtract:

   $$\text{segment area} = \text{sector area} - \text{triangle area}$$

5. How to find the triangle area

In many 9th Grade problems, the triangle inside the sector will be one you can solve using a familiar area formula.

The most common triangle formula is:

$$A = \frac{1}{2}bh$$

If the triangle is made by two radii and forms a special triangle, you may be given enough information to find the base and height, or the area may be provided in the diagram.

Sometimes, when the central angle is \(90^\circ\), the triangle is a right triangle, which makes the area easier to find.

Worked Example 1: Finding the area of a sector

A circle has radius \(8\text{ cm}\). Find the area of a sector with central angle \(45^\circ\).

Step 1: Write the sector formula

$$A = \frac{\theta}{360}\pi r^2$$

Step 2: Substitute the values

$$A = \frac{45}{360}\pi(8^2)$$

$$A = \frac{1}{8}\pi(64)$$

$$A = 8\pi$$

Answer: The area of the sector is \(8\pi\text{ cm}^2\).

Worked Example 2: A larger sector

A circle has radius \(10\text{ m}\). Find the area of a sector with central angle \(144^\circ\).

Step 1: Use the formula

$$A = \frac{\theta}{360}\pi r^2$$

Step 2: Substitute

$$A = \frac{144}{360}\pi(10^2)$$

$$A = \frac{144}{360}\pi(100)$$

Simplify \(\frac{144}{360} = \frac{2}{5}\).

$$A = \frac{2}{5}(100\pi)$$

$$A = 40\pi$$

Answer: The area of the sector is \(40\pi\text{ m}^2\).

Worked Example 3: Finding the area of a segment

A circle has radius \(6\text{ cm}\). A sector has central angle \(90^\circ\). Find the area of the segment.

Step 1: Find the area of the sector

$$A_{\text{sector}} = \frac{90}{360}\pi(6^2)$$

$$A_{\text{sector}} = \frac{1}{4}\pi(36)$$

$$A_{\text{sector}} = 9\pi$$

Step 2: Find the area of the triangle

The two radii are each \(6\text{ cm}\), and since the central angle is \(90^\circ\), they form a right triangle.

So we can use base \(= 6\) and height \(= 6\):

$$A_{\text{triangle}} = \frac{1}{2}bh$$

$$A_{\text{triangle}} = \frac{1}{2}(6)(6) = 18$$

Step 3: Subtract

$$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}}$$

$$A_{\text{segment}} = 9\pi - 18$$

Answer: The area of the segment is \((9\pi - 18)\text{ cm}^2\).

Worked Example 4: Segment with a given triangle area

A circle has radius \(12\text{ cm}\). A sector has central angle \(60^\circ\). The triangle formed inside the sector has area \(36\sqrt{3}\text{ cm}^2\). Find the area of the segment.

Step 1: Find the area of the sector

$$A_{\text{sector}} = \frac{60}{360}\pi(12^2)$$

$$A_{\text{sector}} = \frac{1}{6}\pi(144)$$

$$A_{\text{sector}} = 24\pi$$

Step 2: Subtract the triangle area

$$A_{\text{segment}} = 24\pi - 36\sqrt{3}$$

Answer: The area of the segment is \((24\pi - 36\sqrt{3})\text{ cm}^2\).

6. Important ideas to remember

• A sector is a fraction of a circle.

• A segment is part of a sector after the triangle is removed.

• Always use the central angle in the formula for a sector.

• For segment area, subtract the triangle area from the sector area.

• Check whether your answer should be exact, like \(12\pi\), or rounded to a decimal.

7. Common mistakes

• Using the diameter instead of the radius: In \(\pi r^2\), \(r\) is the radius, not the diameter.

• Forgetting to divide by 360: A sector is only part of a circle.

• Confusing sector and segment: A sector includes the triangle part, but a segment does not.

• Subtracting in the wrong order: For a segment, do sector minus triangle, not triangle minus sector.

8. Quick check

Try these on your own:

1. Find the area of a sector with radius \(9\text{ cm}\) and angle \(120^\circ\).

2. Find the area of a sector with radius \(5\text{ m}\) and angle \(72^\circ\).

3. A circle has radius \(4\text{ cm}\), and a sector has angle \(90^\circ\). Find the area of the segment.

Answers:

1. $$\frac{120}{360}\pi(9^2)=\frac{1}{3}\cdot81\pi=27\pi\text{ cm}^2$$

2. $$\frac{72}{360}\pi(5^2)=\frac{1}{5}\cdot25\pi=5\pi\text{ m}^2$$

3. Sector: $$\frac{90}{360}\pi(4^2)=4\pi$$

   Triangle: $$\frac{1}{2}(4)(4)=8$$

   Segment: $$4\pi-8\text{ cm}^2$$

Summary

To find the area of a sector, take the fraction of the full circle using the central angle:

$$\text{Area of sector} = \frac{\theta}{360}\pi r^2$$

To find the area of a segment, subtract the triangle area from the sector area:

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$

If you remember the difference between a sector and a segment, and carefully use the radius and angle, you can solve these problems step by step.

Central and Inscribed Angles

Circles have many important angle relationships. Two of the most useful are central angles and inscribed angles. Understanding how these angles work helps you solve problems about arcs, chords, and angle measures in a circle.

In this lesson, you will learn what central and inscribed angles are, how they are different, and the rule that connects inscribed angles to the arcs they intercept.

1. Important circle parts

• Center: the point in the middle of the circle.
• Radius: a segment from the center to a point on the circle.
• Arc: a part of the circle's edge.
• Chord: a segment with both endpoints on the circle.

Before studying the angles, remember that an arc is measured in degrees, just like angles. For example, a semicircle measures \(180^\circ\), and a full circle measures \(360^\circ\).

2. What is a central angle?

A central angle is an angle whose vertex is at the center of the circle.

If central angle \(\angle AOB\) is formed by radii \(OA\) and \(OB\), then it intercepts arc \(AB\). The measure of a central angle is equal to the measure of its intercepted arc.

So if

$$m\angle AOB = 80^\circ,$$

then

$$m\overset{\frown}{AB} = 80^\circ.$$

This is one of the simplest circle angle rules:

$$\text{Central angle measure} = \text{intercepted arc measure}$$

3. What is an inscribed angle?

An inscribed angle is an angle whose vertex is on the circle, not at the center.

The sides of an inscribed angle are chords of the circle. An inscribed angle also intercepts an arc.

The key rule is:

$$\text{Inscribed angle measure} = \frac{1}{2}(\text{intercepted arc measure})$$

This means an inscribed angle is always half the measure of the arc it intercepts.

For example, if an inscribed angle intercepts an arc measuring \(100^\circ\), then the angle measures

$$\frac{1}{2}(100^\circ)=50^\circ.$$

4. Comparing central and inscribed angles

If a central angle and an inscribed angle intercept the same arc, then the inscribed angle is half the central angle.

That relationship can be written as

$$m\angle \text{inscribed} = \frac{1}{2}m\angle \text{central}$$

or

$$m\angle \text{central} = 2\,m\angle \text{inscribed}.$$

5. What does “intercepted arc” mean?

The intercepted arc is the arc that lies inside the opening of the angle.

To find the intercepted arc:

• Look at where the sides of the angle touch the circle.
• Find the arc between those two points.
• Use the arc that is inside the angle's opening.

This idea is very important because the angle measure depends on the correct intercepted arc.

6. Main rules to remember

• A central angle has its vertex at the center.
• An inscribed angle has its vertex on the circle.
• The measure of a central angle equals the measure of its intercepted arc.
• The measure of an inscribed angle is half the measure of its intercepted arc.
• If two inscribed angles intercept the same arc, then they are equal.

That last fact is useful. Since both angles are half of the same arc, they must have the same measure.

7. Worked Example 1: Finding an arc from a central angle

A central angle measures \(65^\circ\). What is the measure of its intercepted arc?

Step 1: Use the central angle rule.

$$m\angle \text{central} = m\overset{\frown}{arc}$$

Step 2: Substitute the known value.

$$m\overset{\frown}{arc} = 65^\circ$$

Answer: The intercepted arc measures \(65^\circ\).

8. Worked Example 2: Finding an inscribed angle from an arc

An inscribed angle intercepts an arc that measures \(124^\circ\). Find the angle.

Step 1: Use the inscribed angle rule.

$$m\angle = \frac{1}{2}(m\overset{\frown}{arc})$$

Step 2: Substitute the arc measure.

$$m\angle = \frac{1}{2}(124^\circ)$$

Step 3: Calculate.

$$m\angle = 62^\circ$$

Answer: The inscribed angle measures \(62^\circ\).

9. Worked Example 3: Finding an arc from an inscribed angle

An inscribed angle measures \(38^\circ\). What is the measure of its intercepted arc?

Step 1: Write the relationship.

$$m\angle = \frac{1}{2}(m\overset{\frown}{arc})$$

Step 2: Substitute the angle measure.

$$38^\circ = \frac{1}{2}(m\overset{\frown}{arc})$$

Step 3: Multiply both sides by 2.

$$m\overset{\frown}{arc} = 76^\circ$$

Answer: The intercepted arc measures \(76^\circ\).

10. Worked Example 4: Comparing a central angle and an inscribed angle

A central angle and an inscribed angle intercept the same arc. The central angle measures \(150^\circ\). Find the inscribed angle.

Step 1: Use the relationship between the two angles.

$$m\angle \text{inscribed} = \frac{1}{2}m\angle \text{central}$$

Step 2: Substitute the central angle measure.

$$m\angle \text{inscribed} = \frac{1}{2}(150^\circ)$$

Step 3: Calculate.

$$m\angle \text{inscribed} = 75^\circ$$

Answer: The inscribed angle measures \(75^\circ\).

11. Special case: an inscribed angle in a semicircle

If an inscribed angle intercepts a semicircle, then it intercepts an arc of \(180^\circ\).

Using the inscribed angle rule:

$$m\angle = \frac{1}{2}(180^\circ)=90^\circ$$

So an inscribed angle that intercepts a diameter is always a right angle.

This is a very common fact in circle problems.

12. Common mistakes to avoid

• Mixing up central and inscribed angles: Check where the vertex is. At the center means central. On the circle means inscribed.
• Forgetting the factor of \(\frac{1}{2}\): Inscribed angles are half the intercepted arc, not equal to it.
• Using the wrong arc: Make sure you use the arc inside the opening of the angle.
• Not doubling when needed: If you know the inscribed angle and need the arc, multiply by 2.

13. Quick practice ideas

Try these on your own:

1. A central angle measures \(92^\circ\). Find its arc.
2. An inscribed angle intercepts an arc of \(146^\circ\). Find the angle.
3. An inscribed angle measures \(41^\circ\). Find its intercepted arc.
4. Two inscribed angles intercept the same arc. One angle is \(58^\circ\). What is the other angle?

Answers:

1. \(92^\circ\)
2. \(73^\circ\)
3. \(82^\circ\)
4. \(58^\circ\)

14. Lesson summary

A central angle has its vertex at the center of a circle, and its measure is equal to the measure of its intercepted arc. An inscribed angle has its vertex on the circle, and its measure is half the measure of its intercepted arc.

If a central angle and an inscribed angle intercept the same arc, the central angle is twice the inscribed angle. These relationships are key tools for solving many circle geometry problems.

Tangent and Secant Angle Theorems

When lines meet a circle, they create special angle relationships. Two of the most important are the angles formed by secants and tangents. These theorems help us find missing angles and arcs in circle problems.

In this lesson, you will learn what secants and tangents are, how the angle theorems work, and how to use them step by step.

First, some key vocabulary:

• Circle: the set of all points the same distance from a center point.
• Arc: a part of the circle.
• Secant: a line that cuts the circle at two points.
• Tangent: a line that touches the circle at one point.
• Intercepted arc: the arc that an angle “cuts off” in the circle.

1. Angle formed by two secants inside a circle

If two secants intersect inside a circle, the measure of the angle is half the sum of the intercepted arcs.

In symbols:

$$m\angle = \frac{1}{2}(\text{arc}_1 + \text{arc}_2)$$

This means you add the measures of the two arcs cut off by the angle and its vertical angle, then divide by 2.

Important idea: For angles formed inside the circle, use add, then divide by 2.

2. Angle formed outside a circle by two secants

If two secants intersect outside a circle, the measure of the angle is half the difference of the intercepted arcs.

In symbols:

$$m\angle = \frac{1}{2}(\text{larger arc} - \text{smaller arc})$$

You must subtract the smaller arc from the larger arc first, then divide by 2.

Important idea: For angles formed outside the circle, use subtract, then divide by 2.

3. Angle formed outside a circle by a tangent and a secant

If a tangent and a secant meet outside a circle, the angle measure is also half the difference of the intercepted arcs.

In symbols:

$$m\angle = \frac{1}{2}(\text{larger arc} - \text{smaller arc})$$

This is the same rule as for two secants outside the circle.

4. Angle formed outside a circle by two tangents

If two tangents meet outside a circle, the angle measure is half the difference of the intercepted arcs.

So again:

$$m\angle = \frac{1}{2}(\text{larger arc} - \text{smaller arc})$$

Since the two arcs around a circle add to 360, this is sometimes also written in another way. But for 9th Grade, the safest method is to use the difference of arcs rule.

How to tell which rule to use

1. Look at where the angle’s vertex is.
2. If the vertex is inside the circle, use sum of arcs, then divide by 2.
3. If the vertex is outside the circle, use difference of arcs, then divide by 2.

Common mistake to avoid: Students often mix up inside = add and outside = subtract. A good memory trick is:

• Inside the circle: more “surrounded,” so add.
• Outside the circle: comparing two arcs, so subtract.

Worked Example 1: Two secants intersect inside the circle

Suppose two secants intersect inside a circle. The intercepted arcs measure \(70^\circ\) and \(110^\circ\). Find the angle.

Step 1: Since the vertex is inside the circle, use the sum formula.

$$m\angle = \frac{1}{2}(70 + 110)$$

Step 2: Add the arcs.

$$m\angle = \frac{1}{2}(180)$$

Step 3: Divide by 2.

$$m\angle = 90^\circ$$

Answer: The angle measures \(90^\circ\).

Worked Example 2: Two secants intersect outside the circle

Two secants meet outside a circle. The larger intercepted arc is \(160^\circ\), and the smaller intercepted arc is \(80^\circ\). Find the outside angle.

Step 1: The vertex is outside the circle, so use the difference formula.

$$m\angle = \frac{1}{2}(160 - 80)$$

Step 2: Subtract.

$$m\angle = \frac{1}{2}(80)$$

Step 3: Divide by 2.

$$m\angle = 40^\circ$$

Answer: The angle measures \(40^\circ\).

Worked Example 3: Tangent and secant outside the circle

A tangent and a secant meet outside a circle. The larger intercepted arc measures \(210^\circ\), and the smaller intercepted arc measures \(130^\circ\). Find the angle.

Step 1: Because the angle is outside the circle, use the difference formula.

$$m\angle = \frac{1}{2}(210 - 130)$$

Step 2: Subtract the arcs.

$$m\angle = \frac{1}{2}(80)$$

Step 3: Divide by 2.

$$m\angle = 40^\circ$$

Answer: The angle is \(40^\circ\).

Worked Example 4: Finding a missing arc

Two secants intersect outside a circle and form an angle of \(35^\circ\). The smaller intercepted arc is \(50^\circ\). Find the larger intercepted arc.

Step 1: Use the outside-angle formula.

$$35 = \frac{1}{2}(\text{larger arc} - 50)$$

Step 2: Multiply both sides by 2.

$$70 = \text{larger arc} - 50$$

Step 3: Add 50 to both sides.

$$\text{larger arc} = 120$$

Answer: The larger intercepted arc is \(120^\circ\).

Step-by-step strategy for solving problems

1. Identify the lines: are they secants, tangents, or both?
2. Check where the angle is formed: inside or outside the circle.
3. Find the intercepted arcs.
4. Choose the correct formula:

• Inside angle: $$m\angle = \frac{1}{2}(\text{arc}_1 + \text{arc}_2)$$
• Outside angle: $$m\angle = \frac{1}{2}(\text{larger arc} - \text{smaller arc})$$

5. Solve carefully and check that your answer makes sense.

Quick checks for reasonableness

• An angle measure should not be larger than the arcs used to find it.
• If you are working with an outside angle, make sure you subtract larger minus smaller, not the other way around.
• If your angle is negative, something went wrong.
• If the angle is inside the circle, make sure you added the arcs.

Practice thinking

Ask yourself these questions when solving:

• Is the angle inside or outside the circle?
• Do I need to add or subtract the arcs?
• Did I divide by 2 at the end?

Summary

Tangent and secant angle theorems connect angles with intercepted arcs. If the angle is formed inside the circle, take half the sum of the arcs. If the angle is formed outside the circle, take half the difference of the arcs.

Remember this shortcut:

• Inside = add, then divide by 2
• Outside = subtract, then divide by 2

Once you can identify where the angle is and which arcs it intercepts, these problems become much easier.

Chord and Secant Segment Lengths

Circles have many special patterns. One important pattern involves the lengths of segments formed by chords and secants. These relationships help us find missing lengths without measuring the whole circle directly.

In this lesson, you will learn two main ideas:

• What happens when two chords intersect inside a circle.
• What happens when two secants are drawn from a point outside a circle.

These ideas are sometimes called the Power of a Point rules, but you do not need that name to use them. What matters is understanding the formulas and knowing when to use each one.

First, let’s review the vocabulary.

• Chord: a segment with both endpoints on the circle.
• Secant: a line that cuts through the circle at two points.
• External segment: the part of a secant outside the circle.
• Whole secant: the entire secant from the outside point through the circle.

1. Intersecting Chords Inside a Circle

When two chords intersect inside a circle, each chord is split into two parts. A special relationship connects the lengths of these parts.

If one chord has segments of lengths \(a\) and \(b\), and the other chord has segments of lengths \(c\) and \(d\), then:

$$ab = cd$$

This means the product of the two parts of one chord equals the product of the two parts of the other chord.

It is important to multiply the two pieces of the same chord. Do not add them unless the problem asks for the whole chord.

Example of the setup:

• Chord 1 is split into lengths \(4\) and \(6\).
• Chord 2 is split into lengths \(x\) and \(3\).

Then the equation is:

$$4 \cdot 6 = x \cdot 3$$

2. Two Secants from a Point Outside the Circle

Now consider a different situation. Suppose two secants start from the same point outside the circle. Each secant has:

• an external part, which is outside the circle, and
• a whole length, which includes the outside part and the part inside the circle.

In this case, the relationship is:

$$({\text{external}})({\text{whole}}) = ({\text{external}})({\text{whole}})$$

If one secant has external segment \(a\) and whole length \(b\), and the other has external segment \(c\) and whole length \(d\), then:

$$ab = cd$$

Even though this looks similar to the chord formula, the meaning is different. For secants, you multiply:

• the outside part of a secant by
• the entire secant length.

You do not multiply the outside part by only the inside part.

Example of the setup:

• First secant: external segment \(5\), inside segment \(7\), so whole secant is \(12\).
• Second secant: external segment \(x\), whole secant is \(9\).

The equation is:

$$5(12) = x(9)$$

How to Tell Which Rule to Use

Before solving, ask yourself: Where is the intersection point?

• If the segments intersect inside the circle, use the intersecting chords rule.
• If the secants meet at a point outside the circle, use the secant-secant rule.

This is the most important step. Many mistakes happen because students use the right formula in the wrong situation.

Worked Example 1: Intersecting Chords

Two chords intersect inside a circle. One chord is split into segments of lengths \(3\) and \(8\). The other chord is split into segments of lengths \(x\) and \(4\). Find \(x\).

Step 1: Write the intersecting chords equation.

$$3 \cdot 8 = x \cdot 4$$

Step 2: Multiply.

$$24 = 4x$$

Step 3: Solve for \(x\).

$$x = 6$$

Answer: \(x = 6\)

Worked Example 2: Intersecting Chords with a Whole Chord Given

Two chords intersect inside a circle. One chord has one segment of length \(5\) and the other segment of length \(x\). The second chord has total length \(10\), and one of its segments is \(4\). Find \(x\).

Step 1: Find the missing segment on the second chord.

If the whole chord is \(10\) and one part is \(4\), then the other part is:

$$10 - 4 = 6$$

Step 2: Use the intersecting chords formula.

$$5 \cdot x = 4 \cdot 6$$

Step 3: Simplify.

$$5x = 24$$

Step 4: Solve.

$$x = \frac{24}{5} = 4.8$$

Answer: \(x = 4.8\)

Worked Example 3: Two Secants from an External Point

From a point outside a circle, two secants are drawn. On the first secant, the external segment is \(4\) and the inside segment is \(6\). On the second secant, the external segment is \(x\) and the inside segment is \(10\). Find \(x\).

Step 1: Find each whole secant length.

First secant whole length:

$$4 + 6 = 10$$

Second secant whole length:

$$x + 10$$

Step 2: Use the secant-secant formula.

$$4(10) = x(x + 10)$$

Step 3: Simplify.

$$40 = x^2 + 10x$$

Step 4: Move all terms to one side.

$$x^2 + 10x - 40 = 0$$

Step 5: Solve.

This quadratic does not factor easily, so we use the quadratic formula:

$$x = \frac{-10 \pm \sqrt{10^2 - 4(1)(-40)}}{2}$$ $$x = \frac{-10 \pm \sqrt{100 + 160}}{2}$$ $$x = \frac{-10 \pm \sqrt{260}}{2}$$ $$x = -5 \pm \sqrt{65}$$

A length cannot be negative, so we take the positive value:

$$x = -5 + \sqrt{65}$$

Approximate value:

$$x \approx 3.1$$

Answer: \(x \approx 3.1\)

Worked Example 4: Secants with a Whole Length Given

Two secants are drawn from the same point outside a circle. One secant has an external segment of \(3\) and a whole length of \(15\). The other secant has an external segment of \(5\) and a whole length of \(x\). Find \(x\).

Step 1: Use the secant-secant formula.

$$3(15) = 5x$$

Step 2: Simplify.

$$45 = 5x$$

Step 3: Solve.

$$x = 9$$

Answer: \(x = 9\)

Common Mistakes to Avoid

• Mixing up the rules: chords intersect inside; secants meet outside.
• Using only part of a secant instead of the whole secant: for secants, you must use external \(\times\) whole.
• Forgetting to add segments: if a whole secant is not given, add the external and inside parts.
• Using negative answers for lengths: lengths must be positive.

Quick Problem-Solving Steps

1. Look at the diagram and decide whether it is intersecting chords or two secants from an outside point.
2. Label the known segment lengths carefully.
3. Write the correct equation.
4. Solve for the missing value.
5. Check whether your answer makes sense as a length.

Summary

When two chords intersect inside a circle, multiply the two parts of one chord and set that equal to the product of the two parts of the other chord:

$$ab = cd$$

When two secants are drawn from the same point outside a circle, multiply the external segment by the whole secant for each secant:

$$({\text{external}})({\text{whole}}) = ({\text{external}})({\text{whole}})$$

If you first identify the situation correctly, these problems become much easier. Always pay attention to whether the point of intersection is inside or outside the circle.

Standard Equation of a Circle

A circle is the set of all points in a plane that are the same distance from one fixed point. That fixed point is called the center, and the fixed distance is called the radius.

When we graph circles on the coordinate plane, we want an equation that tells us exactly where the center is and how long the radius is. The most useful form is called the standard equation of a circle.

The standard equation is:

$$ (x-h)^2 + (y-k)^2 = r^2 $$

In this equation:

• (h, k) is the center of the circle
• r is the radius

So if you know the center and radius, you can write the equation. If you know the equation, you can find the center and radius.

Where does this equation come from?

The standard equation comes from the distance formula. Recall that the distance between two points (x_1, y_1) and (x_2, y_2) is:

$$ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$

Now imagine a circle with center (h, k). Any point (x, y) on the circle is exactly r units from the center. Using the distance formula, we get:

$$ r = \sqrt{(x-h)^2 + (y-k)^2} $$

Now square both sides to remove the square root:

$$ r^2 = (x-h)^2 + (y-k)^2 $$

Rewriting gives the standard form:

$$ (x-h)^2 + (y-k)^2 = r^2 $$

This equation works because every point on the circle is the same distance from the center.

How to read the equation

It is very important to notice that the signs inside the parentheses can feel backwards.

• If the equation has (x-3)^2, then h = 3
• If the equation has (x+4)^2, then h = -4
• If the equation has (y-2)^2, then k = 2
• If the equation has (y+5)^2, then k = -5

That happens because (y+5) is the same as (y-(-5)).

Special case: center at the origin

If the center is (0,0), then the equation becomes simpler:

$$ x^2 + y^2 = r^2 $$

For example, a circle centered at the origin with radius 6 has equation:

$$ x^2 + y^2 = 36 $$

How to graph a circle from its equation

1. Find the center (h, k).
2. Find the radius by taking the square root of the number on the right side.
3. Plot the center.
4. From the center, move right, left, up, and down by the radius.
5. Draw a smooth round curve through those points.

For example, if the equation is (x-2)^2 + (y+1)^2 = 9, then:

• Center: (2,-1)
• Radius: \sqrt{9} = 3

From the center (2,-1), go:

• Right 3 to (5,-1)
• Left 3 to (-1,-1)
• Up 3 to (2,2)
• Down 3 to (2,-4)

Then sketch the circle through those four points.

How to write the equation of a circle

If you know the center and radius, use the formula directly:

$$ (x-h)^2 + (y-k)^2 = r^2 $$

Just substitute the values of h, k, and r.

Worked Example 1: Write the equation from center and radius

Write the equation of a circle with center (4,-2) and radius 5.

Step 1: Use the standard form.

$$ (x-h)^2 + (y-k)^2 = r^2 $$

Step 2: Substitute h = 4, k = -2, and r = 5.

$$ (x-4)^2 + (y-(-2))^2 = 5^2 $$

Step 3: Simplify.

$$ (x-4)^2 + (y+2)^2 = 25 $$

Answer: The equation is

$$ (x-4)^2 + (y+2)^2 = 25 $$

Worked Example 2: Find the center and radius from the equation

Find the center and radius of the circle:

$$ (x+3)^2 + (y-1)^2 = 16 $$

Step 1: Compare with standard form.

$$ (x-h)^2 + (y-k)^2 = r^2 $$

Step 2: Identify each part.

• (x+3)^2 = (x-(-3))^2, so h = -3
• (y-1)^2, so k = 1
• r^2 = 16, so r = 4

Answer:

• Center: (-3, 1)
• Radius: 4

Worked Example 3: Graph a circle from its equation

Graph the circle:

$$ (x-1)^2 + (y-2)^2 = 4 $$

Step 1: Find the center.

The center is (1,2).

Step 2: Find the radius.

Since r^2 = 4, the radius is 2.

Step 3: Plot the center (1,2).

Step 4: Find easy points using the radius.

• Right 2: (3,2)
• Left 2: (-1,2)
• Up 2: (1,4)
• Down 2: (1,0)

Step 5: Draw a smooth circle through these points.

Worked Example 4: Write an equation from a center and a point on the circle

A circle has center (2,3) and passes through the point (6,3). Find the equation.

Step 1: Find the radius.

The point (6,3) is 4 units to the right of (2,3), so the radius is 4.

Step 2: Use standard form.

$$ (x-h)^2 + (y-k)^2 = r^2 $$

Step 3: Substitute the values.

$$ (x-2)^2 + (y-3)^2 = 4^2 $$

Step 4: Simplify.

$$ (x-2)^2 + (y-3)^2 = 16 $$

Answer:

$$ (x-2)^2 + (y-3)^2 = 16 $$

Common mistakes to avoid

• Forgetting the sign change in the center. In (x+2)^2, the x-coordinate of the center is -2, not 2.
• Using r instead of r^2. The number on the right side is r^2, so you may need a square root to find the radius.
• Mixing up center and radius. The center is an ordered pair, but the radius is just one positive number.
• Not squaring the radius. If the radius is 7, then the equation uses 49 on the right side.

Quick check questions

1. What is the center and radius of (x-5)^2 + (y+2)^2 = 36?
2. Write the equation of a circle with center (-1,-4) and radius 3.
3. What is the equation of a circle centered at the origin with radius 8?

Answers:

• 1. Center (5,-2), radius 6
• 2. (x+1)^2 + (y+4)^2 = 9
• 3. x^2 + y^2 = 64

Summary

The standard equation of a circle is:

$$ (x-h)^2 + (y-k)^2 = r^2 $$

This equation tells you the circles center (h,k) and radius r. It comes from the distance formula because every point on a circle is the same distance from the center.

To graph a circle, plot the center and move right, left, up, and down by the radius. To write an equation, substitute the center and radius into the standard form carefully, paying close attention to the signs.

Completing the Square for Circle Equations

When a circle equation is written in its standard form, it is easy to see the center and radius.

$$ (x-h)^2+(y-k)^2=r^2 $$

In this form, the center is \\( (h,k) \\) and the radius is \\( r \\).

But sometimes a circle is given in an expanded form, like this:

$$ x^2+y^2+6x-8y+9=0 $$

In that form, the center and radius are not easy to see right away. That is where completing the square helps.

This lesson will show you how to rewrite an expanded circle equation into standard form so you can quickly find the center and radius.

1. Review: What standard form tells us

A circle in standard form looks like:

$$ (x-h)^2+(y-k)^2=r^2 $$

• The center is \\( (h,k) \\).
• The radius is \\( r \\).

Example: In

$$ (x-3)^2+(y+2)^2=16 $$

the center is \\( (3,-2) \\) because \\( y+2=y-(-2) \\), and the radius is \\( 4 \\) because \\( 16=4^2 \\).

2. Why we complete the square

Expanded circle equations usually have \\(x^2\\), \\(y^2\\), an \\(x\\) term, a \\(y\\) term, and a constant.

A common form is:

$$ x^2+y^2+Dx+Ey+F=0 $$

To turn this into standard form, we group the \\(x\\)-terms together and the \\(y\\)-terms together, then complete the square for each group.

3. How to complete the square

To complete the square for an expression like

$$ x^2+bx $$

follow these steps:

1. Take the coefficient of \\(x\\), which is \\(b\\).
2. Divide it by 2.
3. Square the result.

So the number you add is:

$$ \left(\frac{b}{2}\right)^2 $$

Then the trinomial becomes a perfect square:

$$ x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2 $$

For example:

$$ x^2+6x $$

Take half of 6, which is 3, then square it:

$$ 3^2=9 $$

So

$$ x^2+6x+9=(x+3)^2 $$

Another example:

$$ y^2-8y $$

Take half of \\(-8\\), which is \\(-4\\), then square it:

$$ (-4)^2=16 $$

So

$$ y^2-8y+16=(y-4)^2 $$

4. The main process for circle equations

When changing an expanded circle equation into standard form, use this process:

1. Move the constant term to the other side if needed.
2. Group the \\(x\\)-terms and the \\(y\\)-terms.
3. Complete the square for the \\(x\\)-group.
4. Complete the square for the \\(y\\)-group.
5. Add the same numbers to the other side of the equation.
6. Rewrite in standard form.
7. Read the center and radius.

Important: Whatever you add to the left side, you must also balance by adding it to the right side.

Worked Example 1: A basic circle equation

Rewrite in standard form and find the center and radius:

$$ x^2+y^2+6x-8y=0 $$

Step 1: Group the \\(x\\)-terms and \\(y\\)-terms.

$$ (x^2+6x)+(y^2-8y)=0 $$

Step 2: Complete the square for each group.

For \\(x^2+6x\\): half of 6 is 3, and \\(3^2=9\\).

For \\(y^2-8y\\): half of \\(-8\\) is \\(-4\\), and \\((-4)^2=16\\).

Add 9 and 16 to both sides:

$$ (x^2+6x+9)+(y^2-8y+16)=0+9+16 $$ $$ (x+3)^2+(y-4)^2=25 $$

Step 3: Identify the center and radius.

• Center: \\( (-3,4) \\)
• Radius: \\( 5 \\)

Be careful: \\(x+3\\) means \\(x-(-3)\\), so the \\(x\\)-coordinate of the center is \\(-3\\).

Worked Example 2: A constant is on the left side

Rewrite in standard form and find the center and radius:

$$ x^2+y^2-2x+10y-11=0 $$

Step 1: Move the constant term to the right side.

$$ x^2-2x+y^2+10y=11 $$

Step 2: Complete the square for each variable.

For \\(x^2-2x\\): half of \\(-2\\) is \\(-1\\), and \\((-1)^2=1\\).

For \\(y^2+10y\\): half of 10 is 5, and \\(5^2=25\\).

Add 1 and 25 to both sides:

$$ (x^2-2x+1)+(y^2+10y+25)=11+1+25 $$ $$ (x-1)^2+(y+5)^2=37 $$

Step 3: Identify the center and radius.

• Center: \\( (1,-5) \\)
• Radius: \\( \sqrt{37} \\)

The radius does not always come out to a whole number. That is okay.

Worked Example 3: Larger numbers

Rewrite in standard form and find the center and radius:

$$ x^2+y^2+12x-4y-12=0 $$

Step 1: Move the constant term.

$$ x^2+12x+y^2-4y=12 $$

Step 2: Complete the square.

For \\(x^2+12x\\): half of 12 is 6, and \\(6^2=36\\).

For \\(y^2-4y\\): half of \\(-4\\) is \\(-2\\), and \\((-2)^2=4\\).

Add 36 and 4 to both sides:

$$ (x^2+12x+36)+(y^2-4y+4)=12+36+4 $$ $$ (x+6)^2+(y-2)^2=52 $$

Step 3: Identify the center and radius.

• Center: \\( (-6,2) \\)
• Radius: \\( \sqrt{52} \\)

You can leave the radius as \\( \sqrt{52} \\), or simplify it:

$$ \sqrt{52}=\sqrt{4\cdot 13}=2\sqrt{13} $$

So the radius is also \\( 2\sqrt{13} \\).

Worked Example 4: Check if it really makes a circle

Rewrite in standard form and decide whether the equation represents a circle:

$$ x^2+y^2+4x-6y+20=0 $$

Step 1: Move the constant term.

$$ x^2+4x+y^2-6y=-20 $$

Step 2: Complete the square.

For \\(x^2+4x\\): half of 4 is 2, and \\(2^2=4\\).

For \\(y^2-6y\\): half of \\(-6\\) is \\(-3\\), and \\((-3)^2=9\\).

Add 4 and 9 to both sides:

$$ (x^2+4x+4)+(y^2-6y+9)=-20+4+9 $$ $$ (x+2)^2+(y-3)^2=-7 $$

Now look at the right side. It says

$$ r^2=-7 $$

But a radius cannot be a negative number, and \\(r^2\\) cannot be negative.

So this equation does not represent a real circle.

5. Common mistakes to avoid

• Forgetting to move the constant first: It is usually easier to complete the square after the constant is on the other side.
• Adding to only one side: If you add a number on the left, you must add it on the right too.
• Using the wrong sign for the center: In \\( (x-4)^2 \\), the center coordinate is 4. In \\( (x+4)^2 \\), the center coordinate is \\(-4\\).
• Not squaring after dividing by 2: For example, with \\(x^2+8x\\), half of 8 is 4, but the number you add is \\(4^2=16\\), not 4.
• Thinking the right side is the radius: The right side is \\(r^2\\), so you must take the square root to get the radius.

6. Quick pattern to remember

If you have

$$ x^2+y^2+Dx+Ey+F=0 $$

then after completing the square, the standard form will be

$$ \left(x+\frac{D}{2}\right)^2+\left(y+\frac{E}{2}\right)^2=\text{number} $$

Then:

• The center is the opposite of the numbers inside the parentheses.
• The radius is the square root of the number on the right.

7. Practice thinking

Here are some questions you can ask yourself while solving:

• Did I group the \\(x\\)-terms and \\(y\\)-terms correctly?
• Did I take half of each linear coefficient?
• Did I square those halves?
• Did I add the same values to both sides?
• Did I read the signs in the center correctly?
• Did I take the square root to find the radius?

8. Brief summary

Completing the square changes an expanded circle equation into standard form.

Once the equation is written as

$$ (x-h)^2+(y-k)^2=r^2 $$

you can easily identify the center \\( (h,k) \\) and the radius \\( r \\).

The key steps are to group the \\(x\\)-terms and \\(y\\)-terms, add the correct values to complete each square, and keep the equation balanced.

With practice, this process becomes a reliable way to understand circle equations and graph circles correctly.