1. EXAMPLE 1
Graph an equation of a circle
Graph y2 = – x Identify the radius of the circle.
SOLUTION
STEP 1
Rewrite the equation y2 = – x in standard form as x2 + y2 = 36.
STEP 2
Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius
r = = 6.

2. EXAMPLE 1
Graph an equation of a circle
STEP 3
Draw the circle. First plot several convenient points that are 6 units from the origin, such as (0, 6), (6, 0), (0, –6), and (–6, 0). Then draw the circle that passes through the points.

3. Write an equation of a circle
EXAMPLE 2
Write an equation of a circle
The point (2, –5) lies on a circle whose center is the origin. Write the standard form of the equation of the circle.
SOLUTION
Because the point (2, –5) lies on the circle, the circle’s radius r must be the distance between the center (0, 0) and (2, –5). Use the distance formula.
r = (2 – 0)2 + (–5 – 0)2 = =
The radius is 29

4. Write an equation of a circle
EXAMPLE 2
Write an equation of a circle
Use the standard form with r to write an equation of the circle. =
x2 + y2 = r2
Standard form
= ( )2
x2 + y2
Substitute for r 29
x2 + y2 = 29
Simplify

5. EXAMPLE 3
Standardized Test Practice
SOLUTION
A line tangent to a circle is perpendicular to the radius at the point of tangency. Because the radius to the point
(1–3, 2) has slope =
2 – 0
– 3 – 0 = 2 3 – m

6. Standardized Test Practice
EXAMPLE 3
Standardized Test Practice
the slope of the tangent line at (–3, 2) is the negative reciprocal of or An equation of 2 3 2 – 3
the tangent line is as follows:
y – 2 = (x – (– 3)) 3 2
Point-slope form 3 2
y – 2 = x + 9
Distributive property 3 2 13
y =
x +
Solve for y.
ANSWER
The correct answer is C.

7. GUIDED PRACTICE
for Examples 1, 2, and 3
Graph the equation. Identify the radius of the circle. 1.
x2 + y2 = 9
SOLUTION 3

8. GUIDED PRACTICE
for Examples 1, 2, and 3 2.
y2 = –x2 + 49
SOLUTION 7

9. GUIDED PRACTICE
for Examples 1, 2, and 3 3.
x2 – 18 = –y2
SOLUTION 2

10. GUIDED PRACTICE
for Examples 1, 2, and 3
4. Write the standard form of the equation of the circle that passes through (5, –1) and whose center is the origin.
SOLUTION
x2 + y2 = 26
5. Write an equation of the line tangent to the circle x2 + y2 = 37 at (6, 1).
SOLUTION
y = –6x + 37