Circle equations: center, radius, and degenerate cases

Content updated: October 22, 2026

A circle equation follows from requiring a point’s distance from the center to equal a positive radius. Completing the square must distinguish a genuine circle, a single point, and an equation with no real solutions.

On this page: The idea · Method · Examples · Practice · Further learning

1. The equation from the definition

Let the center be $I(a;b)$ and the radius be $R>0$. A point $M(x;y)$ lies on the circle if and only if $IM=R$. Since both sides are nonnegative, squaring is equivalent to

A center and a positive radius therefore determine a unique circle. Keep the signs in $x-a$ and $y-b$ correct: for example, a center with y-coordinate $-2$ produces the term $(y+2)^2$.

2. Expanded form and its conditions

Expanding and setting $c=a^2+b^2-R^2$ gives the equation $x^2+y^2-2ax-2by+c=0$. Conversely, completing the square gives

• If $a^2+b^2-c>0$, this is a circle with center $(a;b)$ and radius $\sqrt{a^2+b^2-c}$.
• If $a^2+b^2-c=0$, its solution set is only the point $(a;b)$, not a nondegenerate circle.
• If $a^2+b^2-c<0$, there are no real solutions, since a sum of two squares is nonnegative.

A single point is sometimes called a degenerate circle of radius $0$; that convention must be stated. This lesson uses “circle” for a positive radius. A negative right side does not produce a real parabola, hyperbola, or ellipse: the real locus is empty.

If the coefficients of $x^2,y^2$ are equal to the same number different from $0$, first divide the entire equation by it. Do not apply this center–radius formula directly when the two squared terms have different coefficients or an $xy$ term is present.

3. Worked examples

Example 1. Expanding from a center and radius

The circle with center $I(1;-2)$ and radius $3$ has equation $(x-1)^2+(y+2)^2=9$. Expanding gives

The constant term is $1+4-9=-4$. As a check, the point $(4;-2)$ is at distance $3$ from the center and satisfies the expanded equation.

Example 2. Reading the center and radius

For $x^2+y^2+2x-6y+6=0$, we have $a=-1$, $b=3$, and $c=6$, so $R^2=1+9-6=4$. Since $R^2>0$, this is a circle with center $I(-1;3)$ and radius $2$. Its center–radius form is $(x+1)^2+(y-3)^2=4$.

Example 3. No real points

The equation $x^2+y^2-4x+2y+6=0$ is equivalent to $(x-2)^2+(y+1)^2=-1$. It has no real points and does not define a real circle. A square root of a negative number cannot serve as a real radius.

Example 4. A single point

The equation $x^2+y^2-2x+4y+5=0$ is equivalent to $(x-1)^2+(y+2)^2=0$. Both squares must equal $0$, so the solution set consists only of $I(1;-2)$. This case must be distinguished from $R>0$.

4. Practice exercises

1. Find the center and radius of $x^2+y^2-6x+8y+16=0$.
2. Classify the solution set of $x^2+y^2+2x-4y+5=0$.
3. Write the equation of the circle with center $I(2;1)$ passing through $A(5;5)$.

5. Common mistakes

• Reading the coefficient $-2a$ as the center’s x-coordinate $-2a$.
• Taking $R^2$ as the radius without taking its positive square root.
• Using the condition $R^2\ge0$ and merging the single-point case with a nondegenerate circle.
• Expanding the constant term incorrectly or failing to check a point on the circle.

6. Related lessons

• Tangent equations of a circle (English lesson)
• Distances in Oxy (English lesson)
• Translating a circle (English lesson)

Original exercise resource on Scribd.