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```					10-2        Circles

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

1
1. (5, 7) and (–1, 6)    6

2. (3, –4) and (–4, 3)   –1

Holt Algebra 2
10-2        Circles

Warm Up
Find the distance between each pair of
points.

3. (–2, 12) and (6, –3)   17

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

Holt Algebra 2
10-2        Circles

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

Holt Algebra 2
10-2        Circles

Vocabulary
circle
tangent

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
Example 1: Using the Distance Formula to Write the
Equation of a Circle
Write the equation of a circle with center (–3, 4)
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
Check It Out! Example 1

Write the equation of a circle with center (4, 2)
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.

Holt Algebra 2
10-2        Circles

Notice 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.

Holt Algebra 2
10-2        Circles

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

Holt Algebra 2
10-2        Circles
Example 2A: Writing the Equation 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

Holt Algebra 2
10-2        Circles
Example 2B: Writing the Equation 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

(x + 4)2 + (y – 11)2 = 152   Substitute the values into the
equation of a circle.

(x + 4)2 + (y – 11)2 = 225
Holt Algebra 2
10-2        Circles
Check It Out! Example 2

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

Use the Distance Formula

(x + 3)2 + (y – 5)2 = 132       Substitute the values into
the equation of a circle.
2      2
(x + 3) + (y – 5) = 169

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
Example 3: Consumer Application
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       2     2
(–2 + 1) + (–3 – 1) < 4
(–1)2 + (–4)2 < 42
1 + 16 < 16 x Point F (–2, –3) is not inside the circle.
Holt Algebra 2
10-2        Circles
Check It Out! 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
2
will satisfy the inequality (x – 2) +
(y + 1)2 < 32. Points C and E are
Check Point B (1, 2) is near the boundary.
2        2     2
(1 – 2) + (2 + 1) < 3
(–1)2 + (3)2 < 32
1+9<9 x           Point B (1, 2) is not inside the circle.
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.

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

Holt Algebra 2
10-2        Circles
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.

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

Holt Algebra 2
10-2        Circles
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 ).
5
The slope of the radius is    2    .

Because the slopes of perpendicular lines are
2
negative reciprocals, the slope of the tangent is –        5
.
Holt Algebra 2
10-2        Circles
Example 4 Continued

Step 3 Find the slope-intercept equation of the
tangent by using the point (2, 5) and the slope
m=– 2.5

Use the point-slope formula.

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

Rewrite in slope-intercept form.

Holt Algebra 2
10-2        Circles
Example 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
Check It Out! 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.

Holt Algebra 2
10-2        Circles
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 ).
–3
The slope of the radius is    4
.

Because the slopes of perpendicular lines are
negative reciprocals, the slope of the tangent is               .
Holt Algebra 2
10-2        Circles
Check It Out! Example 4 Continued

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

Use the point-slope formula.

4
Substitute (5, –5 ) for (x1 , y1 ) and   3
for m.
Rewrite in slope-intercept form.

Holt Algebra 2
10-2        Circles
Check It Out! 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.

Holt Algebra 2

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