• circle
• center
• radius
Write an Equation Given the
Radius
LANDSCAPING The plan for a park puts the
center of a circular pond of radius 0.6 mile at
2.5 miles east and 3.8 miles south of the park
headquarters. Write an equation to represent the
border of the pond, using the headquarters as the
origin.
Since the headquarters is at (0, 0), the center of the
pond is at (2.5, –3.8) with radius 0.6 mile.
(x – h)2 + (y – k)2 = r2
(x – 2.5)2 + (y + 3.8)2 = 0.62
(x – 2.5)2 + (y + 3.8)2 = 0.36
Equation of a circle
(h, k) = (2.5, –3.8),
r = 0.6
Simplify.
Answer: The equation is (x – 2.5)2 + (y + 3.8)2 = 0.36.
LANDSCAPING The plan for a park puts the center
of a circular pond of radius 0.5 mile at 3.5 miles west
and 2.6 miles north of the park headquarters. What
is an equation to represent the border of the pond,
using the headquarters as the origin?
A. (x – 3.5)2 + (y + 2.6)2 = 0.25
C. (x + 3.5)2 + (y – 2.6)2 = 0.25
A
0%
0%
B
D. (x – 3.5)2 + (y + 2.6)2 = 0.5
A.
B.
C.
0%
D.
A
B
C
0%
D
D
2
C
B. (x + 3.5) + (y – 2.6) = 0.5
2
Write an Equation from a Graph
Write an equation for the graph.
(x – h)2 + (y – k)2 = r2 Standard form
center ( h, k)
x = 7, y = 3, h = 0, k = 3
(7 – 0)2 + (3 – 3)2= r2
(7)2 + (0)2=r2
49 + 0=r2
49=r2
Answer: So, the equation of the circle is
(x)2 + (y – 3)2 = 49.
Write an equation for the graph.
A. (x – 14)2 + (y – 6)2 = 200
B. (x – 6)2 + (y)2 = 196
C. (x)2 + (y – 6)2 = 36
D. (x)2 + (y – 6)2 = 196
A.
B.
C.
D.
A
B
C
D
Write an Equation Given a Diameter
Write an equation for a circle if the endpoints of the
diameter are at (2, 8) and (2, –2).
First, find the center of the circle.
Midpoint Formula
(x1, y1) = (2, 8),
(x2, y2) = (2, –2)
Add.
Simplify.
Write an Equation Given a Diameter
Now find the radius.
Distance Formula
(x1, y1) = (2, 3),
(x2, y2) = (2, 8)
Subtract.
Simplify.
The radius of the circle is 5 units, so r2 = 25.
Substitute h, k, and r2 into the standard form of the
equation of a circle.
Write an Equation Given a Diameter
Answer: An equation of the circle is
(x – 2)2 + (y – 3)2 = 25.
Write an equation for a circle if the endpoints of the
diameter are at (3, 5) and (3, –7).
A. (x – 3)2 + (y + 1)2 = 36
A
0%
0%
B
D. (x + 3)2 + (y + 6)2 = 1
A
B
C
0%
D
D
C. (x – 3)2 + (y – 6)2 = 1
A.
B.
C.
0%
D.
C
B. (x + 3)2 + (y – 1)2 = 6
Graph an Equation in Standard Form
Find the center and radius of the circle with
equation x2 + y2 = 16. Then graph the circle.
The center is at (0, 0) and the radius is 4.
Answer: center (0, 0);
r=4
Find the center and radius of the circle with equation
x2 + y2 = 9. Then graph the circle.
A. center (0, 0);
radius = 9
0%
B
0%
A
D. center (3, 0);
radius = 3
A
B
C
0%
D
D
C. center (0, 3);
radius = 9
A.
B.
C.
0%
D.
C
B. center (0, 0);
radius = 3
Graph an Equation Not in Standard Form
Find the center and radius of the circle with
equation x2 + y2 + 6x – 7 = 0. Then graph the circle.
Complete the square.
x2 + y2 + 6x – 7 = 0
x2 + 6x + ■ + y2 = 7 + ■
x2 + 6x + 9 + y2 = 7 + 9
(x + 3)2 + y2 = 16
Answer: The center is at (–3, 0)
and the radius is 4.
Find the center and radius of the circle with equation
x2 + y2 + 8x – 4y + 11 = 0. Then graph the circle.
A. center (–4, 2);
radius = 9
0%
B
0%
A
D. center (4, –2);
radius = 3
A
B
C
0%
D
D
C. center (–4, 2);
radius = 3
A.
B.
C.
0%
D.
C
B. center (4, –2);
radius = 9