Conic Sections: Circles
Example 1
Write the standard form of the equation of the circle that is tangent to the x-axis and has its center
at (-4, 2). Then graph the equation.
Since the circle is tangent to the x-axis, the distance
from the center of the circle to the x-axis is the radius.
The center is 2 units above the x-axis. Therefore, the
radius is 2.
(x - h)
2
+ (y - k)
2
= r
2
Standard form
(x - (-4))
2
+ (y - 2)
2
= 2
2
h = -4, k = 2, r = 2
(x + 4)
2
+ (y - 2)
2
= 4
The standard form of the equation of this circle
is (x + 4)
2
+ (y - 2)
2
= 4.
Example 2
GARDENING A plan has been drawn up to include a circular garden at a new community park.
On the plans, the center of the garden is located at grid location (12, 4) and the radius of the garden
is to be 15 feet.
a. Write an equation that describes the garden.
b. Graph the equation found in part a.
a. We know both the center of the circle, (12, 4) and the radius, 15 feet. Use the equation for the
standard form of a circle to find the equation for the garden.
(x - h)
2
+ (y - k)
2
= r
2
Standard form
(x - 12)
2
+ (y - 4)
2
= 15
2
h = 12, k = 4, r = 15
(x -12)
2
+ (y - 4)
2
= 225
The standard form of the equation of this circle is (x -12)
2
+ (y - 4)
2
= 225.
b.
Example 3
The equation of a circle is x
2
+ y
2
+ 4x – 6y + 4 = 0.
a. Write the standard form of the equation.
b. Find the radius and the coordinates of the center.
c. Graph the equation.
a. x
2
+ y
2
+ 4x – 6y + 4 = 0
(x
2
+ 4x + ?) + (y
2
– 6y + ?) = -4 Group to form perfect square trinomials.
(x
2
+ 4x + 4) + (y
2
– 6y + 9) = -4 + 4 + 9 Complete the square.
(x + 2)
2
+ (y – 3)
2
= 9 Factor the trinomials.
(x + 2)
2
+ (y – 3)
2
= 3
2
b. The center of the circle is located at (-2, 3) and the radius is 3.
c. Plot the center at (-2, 3). The radius is 3.
Example 4
Write the standard form of the equation of the circle that passes through the points (-2, 3), (6, -5),
and (0, 7). Then identify the center and the radius of the circle.
Substitute each ordered pair for (x, y) in x
2
+ y
2
+ Dx + Ey + F = 0, to create a system of equations.
(-2)
2
+ (3)
2
+ D(-2) + E(3) + F = 0 (x, y) = (-2, 3)
(6)
2
+ (-5)
2
+ D(6) + E(-5) + F = 0 (x, y) = (6, -5)
(0)
2
+ (7)
2
+ D(0) + E(7) + F = 0 (x, y) = (0, 7)
Simplify the system of equations. -2D + 3E + F + 13 = 0
6D – 5E + F + 61 = 0
7E + F + 49 = 0
The solution to the system is D = -10, E = -4, and F = -21.
The equation of the circle is x
2
+ y
2
– 10x – 4y – 21 = 0. After completing the square, the standard form is
(x – 5)
2
+ (y – 2)
2
= 50. The center of the circle is (5, 2) and the radius is 50 or 5 2.
