Equation of a Circle

Equation of a Circle

How to solve the equation of a circle problems: formula, proof, examples, and their solutions.

Formula

(x - h)^2 + (y - k)^2 = r^2. (h, k): center of the circle, r: radius of the circle

(x - h)2 + (y - k)2 = r2

(h, k): center of the circle
r: radius of the circle

Proof

Equation of a Circle: Proof of the Formula

The distance between the center (h, k)
and any point on the circle (x, y) is r.

So √(x - h)2 + (y - k)2 = r

Distance formula

Square both sides.

Then (x - h)2 + (y - k)2 = r2.

Example 1

Write an equation of the given circle. 1. center at (5, -1), r = 2.

Center: (5, -1), r = 2

(x - 5)2 + (y - (-1))2 = 22

Example 2

Write an equation of the given circle. 2. center at (4, 0), d = 6.

Center: (4, 0), r = 6/2 = 3

(x - 4)2 + (y - 0)2 = 32

Example 3

Write an equation of the given circle. 3. a circle with a diameter whose endpoints are (-1, 3) and (7, 1).

Briefly draw the given conditions.

The endpoints of the diameter are (-1, 3) and (7, 1).

The midpoint of the diameter (M)
is the center of the circle.

And the bisected diameter is the radius r.

Find the center M.

Midpoint formula

Find the radius r.

Use the distance formula
between the center (3, 2)
and the point on the circle (7, 1).

Center: (3, 2), r = √17

(x - 3)2 + (y - 2)2 = (√17)2 = 17

See the graph of (x - 3)2 + (y - 2)2 = 17.

The center (3, 2) is the midpoint of the diameter.

And the distance between (3, 2) and (7, 1)
is the radius: √17.

You used these properties to find the center (3, 2).

Example 4

For the circle x^2 + y^2 - 4x + 10y + 20 = 0, find the center and the radius of the circle.

Make the standard form of the circle.

First, move the constant term +20
to the right side.

Complete the square for x and y.

Write the additional terms on both sides:
+22, +(-5)2

Completing the square

(x - 2)2 + (y - (-5))2 = 32

So center is (2, -5).
And r = 3.

Example 5

The circle passes through (-1, 0), (-2, 1), and (4, 1). Find an equation of the circle.

Write the equation of the circle as
x2 + y2 + Ax + By + C = 0

The goal is to find A, B, and C.

There are 3 variables [A, B, C]
and 3 conditions [(-1, 0), (-2, 1), (4, 1)].

To solve this,
use each condition to remove each variable.

Solving system of equations (3 variables)

First, use (-1, 0) condition:
Put (-1, 0)
into x2 + y2 + Ax + By + C = 0.

Then C = A - 1.

Substitution method

Put C = A - 1
into x2 + y2 + Ax + By + C = 0.

Then x2 + y2 + Ax + By + (A - 1) = 0.

By using (-1, 0) condition, C is removed.

So there are 2 variables [A, B]
and 2 conditions [(-2, 1), (4, 1)].

Next, use (-2, 1) condition:
Put (-2, 1) into
x2 + y2 + Ax + By + (A - 1) = 0.

Then B = A - 4.

Put B = A - 4
into x2 + y2 + Ax + By + (A - 1) = 0.

Then x2 + y2 + Ax + (A - 4)y + (A - 1) = 0.

By using (-2, 1) condition, B is removed.

So there's only 1 variable [A]
and 1 condition [(4, 1)].

Use the final condition (4, 1):
Put (4, 1) into
x2 + y2 + Ax + (A - 4)y + (A - 1) = 0.

Then A = -2.

Put A = -2
into B = A - 4.

Then B = -6.

Put A = -2
into C = A - 1.

Then C = -3.

Put A = -2, B = -6, and C = -3
into x2 + y2 + Ax + By + C = 0.

Then x2 + y2 - 2x - 6y - 3 = 0.

The standard form of the equation is
(x - 1)2 + (y - 3)2 = 13.