# Equation of a Circle

How to write the equation of a circle from the given conditions: formula, examples, and their solutions.

## Formula

(*x* - *h*)^{2} + (*y* - *k*)^{2} = *r*^{2}

(*h*, *k*): Center of the circle*r*: Radius of the circle

## Example 1

The center of the circle is (5, -1).

And its radius is 2.

Then the equation of the circle is

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

-(-1) = +1

2^{2} = 4

So the answer is

(*x* - 5)^{2} + (*y* + 1)^{2} = 4.

When writing the equation of the circle,

you don't have to expand (*x* - 5)^{2} and (*y* + 1)^{2}.

So the given circle looks like this.

Its center is (5, -1).

And its radius is 2.

## Example 2

The center of the circle is (5, -1).*d* = 6

So the radius is, 6/2, 3.

Then the equation of the circle is

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

3^{2} = 9

So the answer is

(*x* - 4)^{2} + *y*^{2} = 9.

So the given circle looks like this.

Its center is (4, 0).

And its radius is 3.

## Example 3

Roughly draw the circle

by using the given conditions.

The endpoints of the diameter are

(-1, 3) and (7, 1).

Set the center of the circle as *M*.

Then the center bisects the diameter.

And the bisected segments are both *r*.

The center *M* is the midpoint of the diameter.

So *M*([-1 + 7]/2, [3 + 1]/2).

Midpoint formula - on a coordinate plane

-1 + 7 = 6

3 + 1 = 4

6/2 = 3

4/2 = 2

So *M*(3, 2).

The radius *r* is the distance between*M*(3, 2) and (7, 1).

So *r* = √(7 - 3)^{2} + (1 - 2)^{2}.

Distance formula

(You can also use the distance between*M*(3, 2) and (-1, 3).)

7 - 3 = 4

1 - 2 = -1

4^{2} = 16

(-1)^{2} = 1

16 + 1 = 17

So *r* = √17.

*M*(3, 2)*r* = √17

So the equation of the circle is

(*x* - 3)^{2} + (*y* - 2)^{2} = (√17)^{2}.

(√17)^{2} = 17

So the answer is

(*x* - 3)^{2} + (*y* - 2)^{2} = 17.

## Example 4

The given circle is in general form:*x*^{2} + *y*^{2} + *Ax* + *By* + *C* = 0.

(*A*, *B*, *C*: Constants)

To find the center and the radius,

change this equation into standard form:

(*x* - *h*)^{2} + (*y* - *k*)^{2} = *r*^{2}.

Move +20 to the right side.

Make two squares using [*x*^{2} - 4*x*] and [*y*^{2} + 10*y*].*x*^{2} - 4*x*

= *x*^{2} - 2⋅2⋅*x*

So, to make a perfect square,

write +2^{2} on both sides.*y*^{2} + 10*y*

= *y*^{2} + 2⋅5⋅*y*

So, to make a perfect square,

write +5^{2} on both sides.

Left side:*x*^{2} - 2⋅2⋅*x* + 2^{2} = (*x* - 2)^{2}*y*^{2} + 2⋅5⋅*y* + 5^{2} = (*y* + 5)^{2}

Factor a perfect square trinomial (*a*^{2} ± 2*ab* + *b*^{2})

Right side:

2^{2} = 4

5^{2} = 25

Left side:

To see the center clearly,

change (*y* + 5)^{2} to (*y* - (-5))^{2}.

Right side:

-20 + 4 + 25 = 9

9 = 3^{2}

So the equation of the circle

in standard form is

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

So the center is (2, -5).

And the radius is 3.

So the given circle looks like this.

Its center is (2, -5).

And its radius is 3.

## Example 5

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

The goal is the find *A*, *B*, and *C*,

by using (-1, 0), (-2, 1), and (4, 1).

Put (-1, 0)

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

Then *A* - *C* = 1.

Put (-2, 1)

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

Then 2*A* - *B* - *C* = 5.

Put (4, 1)

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

Then 4*A* + *B* + *C* = -17.

Then you got this system of linear equations.

(3 equations, 3 variables)*A* - *C* = 1

2*A* - *B* - *C* = 5

4*A* + *B* + *C* = -17

Solve this system to find the variables.

System of equations (3 variables)

Choose these two equations:

2*A* - *B* - *C* = 5*A* - *C* = 1.

Eliminate *C*.

Then *A* - *B* = 4.

Choose these two equations:

4*A* + *B* + *C* = -17*A* - *C* = 1.

Eliminate the same *C*.

Then 5*A* + *B* = -16.

Then you got this system of linear equations.

(2 equations, 2 variables)*A* - *B* = 4

5*A* + *B* = -16

Solve this system to find the variables.

Use the elimination method.

And eliminate *B*.

Then *A* = -2.

Substitution method

Put *A* = -2

into *A* - *B* = 4.

Then *B* = -6.

(You can also put *A* = -2

into 5*A* + *B* = -16:

the other [green equation].)

Put *A* = -2 (and *B* = -6)

into *A* - *C* = 1.

Then *C* = -3.

(You can also put *A* = -2 and *B* = -6

into 2*A* - *B* - *C* = 5

or 4*A* + *B* + *C* = -17:

the other [blue equations].)

*A* = -2*B* = -6*C* = -3

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

Then the equation of the circle is*x*^{2} + *y*^{2} - 2*x* - 6*y* - 3 = 0.