Equations of Circles

Standard form and signs

A circle with centre $(h,k)$ and radius $r>0$ has equation

For $(x-3)^2+(y+2)^2=25$, the centre is $(3,-2)$ and the radius is 5. The sign inside each bracket is part of a subtraction from the coordinate: $y+2=y-(-2)$.

The right-hand side is the squared radius, not the radius. If it is 13, the radius is $\sqrt{13}$. If it is zero, the solution set is a single point. If it is negative, there are no real points because a sum of real squares cannot be negative.

Convert an expanded equation

Rewrite

Group and complete each square

Move the constant and group the coordinate terms:

For the $x$ group, half of $-6$ is $-3$, whose square is 9. For the $y$ group, half of 4 is 2, whose square is 4. Add both quantities to both sides:

Hence

The centre is $(3,-2)$ and the radius is 5. Expand the completed squares to check that the original equation is recovered.

Normalize a common coefficient first

For $2x^2+2y^2-12x+8y-24=0$, divide the entire equation by 2 first. This returns the same example. Completing a square inside a group with a non-unit coefficient without accounting for that coefficient produces an incorrect constant.

A general centre-and-radius test

For an equation already normalized as

completion gives

The centre is $(-D/2,-E/2)$. A real circle exists only when the expression on the right is positive. Equal coefficients on $x^2$ and $y^2$ alone are therefore not enough to guarantee a real circle.

This square-completion approach agrees with the circle and ellipse forms in OpenStax.

Construct a circle from a diameter

Let the diameter endpoints be $A=(-1,2)$ and $B=(5,6)$. The centre is their midpoint:

The squared radius is the squared distance from $C$ to $B$:

Therefore the equation is

Both endpoints give 13 when substituted. Using the full diameter length as the radius would incorrectly make the right-hand side four times as large.

Construct a circle through three points

Three distinct non-collinear points determine a unique circle. To find the circle through $(0,0)$, $(4,0)$ and $(0,2)$, use $x^2+y^2+Dx+Ey+F=0$.

Substituting $(0,0)$ gives $F=0$. Substituting $(4,0)$ gives $16+4D=0$, so $D=-4$. Substituting $(0,2)$ gives $4+2E=0$, so $E=-2$.

Thus

The centre is $(2,1)$ and radius $\sqrt5$. Geometrically, the same centre is where the perpendicular bisectors of the chords meet. Three distinct collinear points cannot all lie on a finite-radius circle.

Solve intersections with a line

Find where $y=x+1$ meets $x^2+y^2=25$. Substitute the line expression into the circle:

Divide by 2 and factor:

The inputs are $x=-4$ and $x=3$. Use the line equation to recover the corresponding outputs: $y=-3$ and $y=4$. The intersection points are therefore $(-4,-3)$ and $(3,4)$.

Both points satisfy both original equations. Finding only the $x$-values would leave the geometric answer incomplete. A repeated quadratic root corresponds to tangency; no real roots means no real intersections.

Related topics

Parabolas introduces a different distance condition. Conic Sections compares circle equations with the other second-degree curves and explains why not every quadratic equation describes a real non-degenerate curve.