# 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 = r2

(h, k): center of the circle

## Proof

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

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

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

## Example 2

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

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

## Example 3

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

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)

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

## Example 4

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

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.