Holt Algebra 2

10-2 Circles10-2 Circles

Holt Algebra2

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Holt Algebra 2

10-2 CirclesWarm UpFind the slope of the line that connects each pair of points.

1. (5, 7) and (–1, 6)

2. (3, –4) and (–4, 3)

Holt Algebra 2

10-2 Circles

Warm UpFind the distance between each pair of points.

3. (–2, 12) and (6, –3)

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

Holt Algebra 2

10-2 Circles

Write an equation for a circle. Graph a circle, and identify its center and radius.

Objectives

Holt Algebra 2

10-2 Circles

circletangent

Vocabulary

Holt Algebra 2

10-2 Circles

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.

Holt Algebra 2

10-2 Circles

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

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

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.

Holt Algebra 2

10-2 Circles

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

Check It Out! Example 1

Holt Algebra 2

10-2 CirclesNotice that r2 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.

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

Helpful Hint

Holt Algebra 2

10-2 Circles

Write the equation of the circle.

Example 2A: Writing the Equation of a Circle

(x – 0)2 + (y – 6)2 = 12

x2 + (y – 6)2 = 1

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

(x – h)2 + (y – k)2 = r2 Equation of a circle

Substitute.

Holt Algebra 2

10-2 Circles

Use the Distance Formula to find the radius.

Substitute the values into the equation of a circle.

(x + 4)2 + (y – 11)2 = 225

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

(x + 4)2 + (y – 11)2 = 152

Write the equation of the circle.

Example 2B: Writing the Equation of a Circle

Holt Algebra 2

10-2 Circles

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

Check It Out! Example 2

Holt Algebra 2

10-2 Circles

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 + (y – k)2 < r2.

The points outside the circle satisfy the inequality (x – h)2 + (y – k)2 > r2.

Holt Algebra 2

10-2 Circles

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)?

Example 3: Consumer Application

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 .

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

Check Point F(–2, –3) is near the boundary.(–2 + 1)2 + (–3 – 1)2 < 42

(–1)2 + (–4)2 < 42

1 + 16 < 16 x

Holt Algebra 2

10-2 Circles

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

Check It Out! Example 3

Holt Algebra 2

10-2 Circles

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.

To review linear functions, see Lesson 2-4.Remember!

Holt Algebra 2

10-2 Circles

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

Example 4: Writing the Equation of a Tangent

Step 1 Identify the center and radius of the circle.

From the equation x2 + y2 = 29, the circle has center of (0, 0) and radius r = .

Holt Algebra 2

10-2 CirclesExample 4 Continued

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

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

Use the slope formula.

The slope of the radius is . 5 2

Because the slopes of perpendicular lines are negative reciprocals, the slope of the tangent is . 2

5 –

Holt Algebra 2

10-2 CirclesExample 4 Continued

Use the point-slope formula.

Rewrite in slope-intercept form.

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

Step 3 Find the slope-intercept equation of the tangent by using the point (2, 5) and the slope m = . 2

5 –

Holt Algebra 2

10-2 CirclesExample 4 Continued

The equation of the line that is tangent to x2 + y2 = 29 at (2, 5) is .

Check Graph the circle and the line.

Holt Algebra 2

10-2 Circles

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

Check It Out! Example 4

Check Graph the circle and the line.