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6

Warm Up

Find the slope of the line that connects each pair of points.

1. (5, 7) and (–1, 6)

2. (3, –4) and (–4, 3)

–1

Find the distance between each pair of points.

3. (–2, 12) and (6, –3)

17

5

4. (1, 5) and (4, 1)

A circle is the set of points in a plane that are a fixed distance, called the radius, from a fixed point, called the center. Because all of the points on a circle are the same distance from the center of the circle, you can use the Distance Formula to find the equation of a circle.

Example 1: Using the Distance Formula to Write the Equation of a Circle

Write the equation of a circle with center (–3, 4) and radius r = 6.

Use the Distance Formula with (x2, y2) = (x, y), (x1, y1) = (–3, 4), and distance equal to the radius, 6.

Use the Distance Formula.

Substitute.

Square both sides.

Check It Out! of a Circle Example 1

Write the equation of a circle with center (4, 2) and radius r = 7.

Use the Distance Formula with (x2, y2) = (x, y), (x1, y1) = (4, 2), and distance equal to the radius, 7.

Use the Distance Formula.

Substitute.

Square both sides.

Notice that of a Circler2 and the center are visible in the equation of a circle. This leads to a general formula for a circle with center (h, k) and radius r.

Helpful Hint of a Circle

If the center of the circle is at the origin, the equation simplifies to x2 + y2 = r2.

Example 2A: Writing the Equation of a Circle of a Circle

Write the equation of the circle.

the circle with center (0, 6) and radius r = 1

(x – h)2 + (y – k)2 = r2

Equation of a circle

(x – 0)2 + (y – 6)2 = 12

Substitute.

x2 + (y – 6)2 = 1

Example 2B: Writing the Equation of a Circle of a Circle

Write the equation of the circle.

the circle with center (–4, 11) and containing the point (5, –1)

Use the Distance Formula to find the radius.

(x + 4)2 + (y – 11)2 = 152

Substitute the values into the equation of a circle.

(x + 4)2 + (y – 11)2 = 225

Check It Out! of a Circle Example 2

Find the equation of the circle with center (–3, 5) and containing the point (9, 10).

Use the Distance Formula to find the radius.

Substitute the values into the equation of a circle.

(x + 3)2 + (y – 5)2 = 132

(x + 3)2 + (y – 5)2 = 169

The location of points in relation to a circle can be described by inequalities. The points inside the circle satisfy the inequality (x – h)2 + (x – k)2 < r2. The points outside the circle satisfy the inequality (x – h)2 + (x – k)2 > r2.

Example 3: Consumer Application described by inequalities. The points inside the circle satisfy the inequality (

Use the map and information given in Example 3 on page 730. Which homes are within 4 miles of a restaurant located at (–1, 1)?

The circle has a center (–1, 1) and radius 4. The points insides the circle will satisfy the inequality (x + 1)2 + (y – 1)2 < 42. Points B, C, D and E are within a 4-mile radius .

Check Point F(–2, –3) is near the boundary.

(–2 + 1)2 + (–3 – 1)2 < 42

(–1)2 + (–4)2 < 42

x

Point F (–2, –3) is not inside the circle.

1 + 16 < 16

Check It Out! described by inequalities. The points inside the circle satisfy the inequality ( Example 3

What if…? Which homes are within a 3-mile radius of a restaurant located at (2, –1)?

The circle has a center (2, –1) and radius 3. The points inside the circle will satisfy the inequality (x – 2)2 + (y + 1)2 < 32. Points C and E are within a 3-mile radius .

Check Point B (1, 2) is near the boundary.

(1 – 2)2 + (2 + 1)2 < 32

(–1)2 + (3)2 < 32

x

1 + 9 < 9

Point B (1, 2) is not inside the circle.

Remember! described by inequalities. The points inside the circle satisfy the inequality (

To review linear functions, see Lesson 2-4.

A tangent is a line in the same plane as the circle that intersects the circle at exactly one point. Recall from geometry that a tangent to a circle is perpendicular to the radius at the point of tangency.

From the equation described by inequalities. The points inside the circle satisfy the inequality ( x2 + y2 = 29, the circle has center of (0, 0) and radius r = .

Example 4: Writing the Equation of a Tangent

Write the equation of the line tangent to the circle x2 + y2 = 29 at the point (2, 5).

Step 1 Identify the center and radius of the circle.

2 described by inequalities. The points inside the circle satisfy the inequality (

–

5

5

2

The slope of the radius is .

Because the slopes of perpendicular lines are negative reciprocals, the slope of the tangent is .

Example 4 Continued

Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent.

Use the slope formula.

Substitute (2, 5) for (x2 , y2 ) and (0, 0) for (x1 , y1 ).

Step 3 described by inequalities. The points inside the circle satisfy the inequality (Find the slope-intercept equation of the tangent by using the point (2, 5) and the slope m = .

2

–

5

Substitute (2, 5) (x1 , y1 ) and – for m.

2

5

Example 4 Continued

Use the point-slope formula.

Rewrite in slope-intercept form.

The equation of the line that is tangent to described by inequalities. The points inside the circle satisfy the inequality ( x2 + y2 = 29 at (2, 5) is .

Example 4 Continued

Check Graph the circle and the line.

Check It Out! described by inequalities. The points inside the circle satisfy the inequality ( Example 4

Write the equation of the line that is tangent to the circle 25 = (x –1)2 + (y + 2)2, at the point (1, –2).

Step 1 Identify the center and radius of the circle.

From the equation 25 = (x – 1)2 +(y + 2)2, the circle has center of (1, –2) and radius r = 5.

– described by inequalities. The points inside the circle satisfy the inequality (3

4

The slope of the radius is .

Because the slopes of perpendicular lines are negative reciprocals, the slope of the tangent is .

Check It Out! Example 4 Continued

Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent.

Use the slope formula.

Substitute (5, –5) for (x2 , y2 ) and (1, –2) for (x1 , y1 ).

Step 3. described by inequalities. The points inside the circle satisfy the inequality (Find the slope-intercept equation of the tangent by using the point (5, –5) and the slope .

Substitute (5, –5 ) for (x1 , y1 ) and for m.

4

3

Check It Out! Example 4 Continued

Use the point-slope formula.

Rewrite in slope-intercept form.

Check It Out! described by inequalities. The points inside the circle satisfy the inequality ( Example 4 Continued

The equation of the line that is tangent to 25 = (x – 1)2 + (y + 2)2 at (5, –5) is .

Check Graph the circle and the line.

Lesson Quiz: Part I described by inequalities. The points inside the circle satisfy the inequality (

1. Write an equation for the circle with center (1, –5) and a radius of .

(x – 1)2 + (y + 5)2 = 10

2. Write an equation for the circle with center (–4, 4) and containing the point (–1, 16).

(x + 4)2 + (y – 4)2 = 153

Lesson Quiz: Part II described by inequalities. The points inside the circle satisfy the inequality (

3. Which points on the graph shown are within 2 units of the point (0, –2.5)?

C, F

Lesson Quiz: Part III described by inequalities. The points inside the circle satisfy the inequality (

4. Write an equation for the line tangent to the circle x2 + y2 = 17 at the point (4, 1).

y – 1 = –4(x – 4)

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