# 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

## 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} = *r*^{2}.

## Example 1

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

(*x* - 5)^{2} + (*y* - (-1))^{2} = 2^{2}

## Example 2

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

(*x* - 4)^{2} + (*y* - 0)^{2} = 3^{2}

## 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

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

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:

+2^{2}, +(-5)^{2}

Completing the square

(*x* - 2)^{2} + (*y* - (-5))^{2} = 3^{2}

So center is (2, -5).

And *r* = 3.

## Example 5

Write the equation of the circle as*x*^{2} + *y*^{2} + *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 *x*^{2} + *y*^{2} + *Ax* + *By* + *C* = 0.

Then *C* = *A* - 1.

Substitution method

Put *C* = *A* - 1

into *x*^{2} + *y*^{2} + *Ax* + *By* + *C* = 0.

Then *x*^{2} + *y*^{2} + *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*x*^{2} + *y*^{2} + *Ax* + *By* + (*A* - 1) = 0.

Then *B* = *A* - 4.

Put *B* = *A* - 4

into *x*^{2} + *y*^{2} + *Ax* + *By* + (*A* - 1) = 0.

Then *x*^{2} + *y*^{2} + *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*x*^{2} + *y*^{2} + *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 *x*^{2} + *y*^{2} + *Ax* + *By* + *C* = 0.

Then *x*^{2} + *y*^{2} - 2*x* - 6*y* - 3 = 0.

The standard form of the equation is

(*x* - 1)^{2} + (*y* - 3)^{2} = 13.