Equation of a Circle

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

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

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

Example 1

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

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 = 22

-(-1) = +1
22 = 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

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

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 = 32

32 = 9

So the answer is
(x - 4)2 + y2 = 9.

So the given circle looks like this.

Its center is (4, 0).
And its radius is 3.

Example 3

Write an equation of the given circle. Endpoints of the diameter: (-1, 3), (7, 1)

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

42 = 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

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

The given circle is in general form:

x2 + y2 + 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 = r2.

Move +20 to the right side.

Make two squares using [x2 - 4x] and [y2 + 10y].

x2 - 4x
= x2 - 2⋅2⋅x
So, to make a perfect square,
write +22 on both sides.

y2 + 10y
= y2 + 2⋅5⋅y
So, to make a perfect square,
write +52 on both sides.

Left side:

x2 - 2⋅2⋅x + 22 = (x - 2)2
y2 + 2⋅5⋅y + 52 = (y + 5)2

Factor a perfect square trinomial (a2 ± 2ab + b2)

Right side:

22 = 4
52 = 25

Left side:

To see the center clearly,
change (y + 5)2 to (y - (-5))2.

Right side:
-20 + 4 + 25 = 9

9 = 32

So the equation of the circle
in standard form is
(x - 2)2 + (y - (-5))2 = 32.

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

A circle passes through the given points. Find an equation of the given circle. (-1, 0), (-2, 1), (4, 1).

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

Then A - C = 1.

Put (-2, 1)
into x2 + y2 + Ax + By + C = 0.

Then 2A - B - C = 5.

Put (4, 1)
into x2 + y2 + Ax + By + C = 0.

Then 4A + B + C = -17.

Then you got this system of linear equations.
(3 equations, 3 variables)

A - C = 1
2A - B - C = 5
4A + B + C = -17

Solve this system to find the variables.

System of equations (3 variables)

Choose these two equations:

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

Eliminate C.

Then A - B = 4.

Choose these two equations:

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

Eliminate the same C.

Then 5A + B = -16.

Then you got this system of linear equations.
(2 equations, 2 variables)

A - B = 4
5A + 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 5A + 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 2A - B - C = 5
or 4A + B + C = -17:
the other [blue equations].)

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

Put these into x2 + y2 + Ax + By + C = 0.

Then the equation of the circle is
x2 + y2 - 2x - 6y - 3 = 0.