The Equation of a Circle
Derive the standard form (x−h)² + (y−k)² = r² from the distance formula. Write equations given center and radius, graph circles, and convert from general form to standard form by completing the square.
A lesson of Geometry, in Circles, about 10 minutes.
By the end of this lesson, you will be able to write the equation of a circle in standard form $(x-h)^2 + (y-k)^2 = r^2$, convert from general form by completing the square, identify the center and radius, and graph circles on the coordinate plane.
The Big Idea
The equation of a circle is the Pythagorean theorem in disguise. Every point $(x, y)$ on the circle is exactly $r$ away from the center $(h, k)$ — and the distance formula gives us the equation directly. Converting from standard to general form (and back) requires completing the square, connecting coordinate algebra to circle geometry.
Standard Form of a Circle
A circle with center $(h, k)$ and radius $r$:
$$(x - h)^2 + (y - k)^2 = r^2$$
Derivation: Distance from $(x, y)$ to center $(h, k)$ equals $r$:
$$\sqrt{(x-h)^2 + (y-k)^2} = r \implies (x-h)^2 + (y-k)^2 = r^2$$
Special case: Circle centered at the origin: $x^2 + y^2 = r^2$
Example 1 — Writing the Equation
Write the equation of a circle with center $(3, -5)$ and radius $7$.
$$(x - 3)^2 + (y + 5)^2 = 49$$
Example 2 — Identifying Center and Radius
Identify the center and radius: $(x + 4)^2 + (y - 2)^2 = 25$.
Center: $(-4, 2)$ (negate the signs inside the parentheses)
Radius: $r = \sqrt{25} = 5$
Example 3 — Writing from Center and a Point
Find the equation of the circle with center $(-1, 3)$ passing through $(5, 3)$.
Radius: $r = |5 - (-1)| = 6$ (same $y$-coordinate, so just horizontal distance)
$$(x + 1)^2 + (y - 3)^2 = 36$$
General case: $r = \sqrt{(5-(-1))^2 + (3-3)^2} = \sqrt{36} = 6$
General Form and Completing the Square
General form: $x^2 + y^2 + Dx + Ey + F = 0$
To convert to standard form, complete the square for $x$ and $y$ separately.
Example 4 — Completing the Square
Convert $x^2 + y^2 + 6x - 4y - 3 = 0$ to standard form.
Group and add completing-the-square terms:
$$(x^2 + 6x) + (y^2 - 4y) = 3$$
$$(x^2 + 6x + 9) + (y^2 - 4y + 4) = 3 + 9 + 4$$
$$(x + 3)^2 + (y - 2)^2 = 16$$
Center: $(-3, 2)$; Radius: $4$
Example 5 — Is the Point on, Inside, or Outside the Circle?
Circle: $(x - 2)^2 + (y + 1)^2 = 25$. Classify the point $(5, 3)$.
Substitute:
$(5-2)^2 + (3+1)^2 = 9 + 16 = 25$
$25 = 25$ → the point is on the circle.