Equation of a Circle

A circle is all points at distance $r$ from a center $(h, k)$. The distance formula from Distance and Midpoint Formulas turns that sentence into an equation.

Why it works

A point $(x, y)$ is on the circle exactly when its distance from $(h, k)$ is $r$: $\sqrt{(x - h)^2 + (y - k)^2} = r$. Both sides are positive, so squaring keeps the same points: $(x - h)^2 + (y - k)^2 = r^2$.

Key idea

$$(x - h)^2 + (y - k)^2 = r^2$$ is the circle with center $(h, k)$ and radius $r$. The signs inside the parentheses are the opposite of the center's coordinates, and the right side is $r^2$, not $r$.

The circle with center $C(1, -1)$ through $P$. Drag $P$: the radius is the distance $CP$, and $(x - 1)^2 + (y + 1)^2 = CP^2$.

Worked example: Writing an equation

Write the equation of the circle with center $(-3, 2)$ and radius $5$.

$h = -3$ and $k = 2$: $(x - (-3))^2 + (y - 2)^2 = 5^2$.
$(x + 3)^2 + (y - 2)^2 = 25$.

Answer: $(x + 3)^2 + (y - 2)^2 = 25$

Worked example: Reading the center and radius

Find the center and radius of $(x - 4)^2 + (y + 1)^2 = 18$.

$x - 4$ means $h = 4$; $y + 1 = y - (-1)$ means $k = -1$. Center $(4, -1)$.
$r^2 = 18$, so $r = \sqrt{18} = 3\sqrt{2}$.

Answer: Center $(4, -1)$, radius $3\sqrt{2}$

Worked example: Through a point

A circle has center $(2, -3)$ and passes through $(5, 1)$. Write its equation.

$r^2 = (5 - 2)^2 + (1 - (-3))^2 = 9 + 16 = 25$.
$(x - 2)^2 + (y + 3)^2 = 25$.

Answer: $(x - 2)^2 + (y + 3)^2 = 25$

Inside, on, or outside? The left side of the equation is the squared distance from the center. So for any point, substitute it into the left side: if the result is less than $r^2$ the point is inside the circle, if it equals $r^2$ it is on the circle, and if it is greater the point is outside.

Worked example: Testing a point

Is $(4, 2)$ inside, on, or outside the circle $(x - 1)^2 + (y + 2)^2 = 20$?

Substitute: $(4 - 1)^2 + (2 + 2)^2 = 9 + 16 = 25$.
$25 > 20$, so the point is farther from the center than the radius.

Answer: Outside

Common mistake

Reading the center of $(x + 2)^2 + (y - 1)^2 = 9$ as $(2, -1)$, or the radius as $9$. The form subtracts the center, so $x + 2$ means $h = -2$: the center is $(-2, 1)$. The right side is $r^2$, so $r = 3$. Check by plugging the center into the left side: you should get 0.

Equation of a Circle

Practice

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