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Equations and Properties of Circles

Find circle centers and radii, distinguish a circle from a single point or an empty set, and test tangency using distance formulas. Three worked examples.

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Equations and Properties of Circles

A circle's equation connects its center and radius with its geometric properties. This tutorial covers the standard and general forms, then uses distance formulas to locate points and lines relative to a circle.

1. Definition and standard form

A circle is the set of points in a plane at a fixed positive distance from a given point. The given point is the center, and the distance is the radius.

For center $(a,b)$ and radius $r>0$, the distance formula gives

$(x-a)^2+(y-b)^2=r^2.$

If the center is the origin, this reduces to $x^2+y^2=r^2$.

2. General form: three possible cases

Expanding the standard form gives

$x^2+y^2+Dx+Ey+F=0.$

Completing the square gives

$\left(x+\frac D2\right)^2+\left(y+\frac E2\right)^2=\frac{D^2+E^2-4F}{4}.$

The left side is a sum of squares of real numbers, so first check the sign of the right side:

In the first case, the center is $(-D/2,-E/2)$ and the radius is

$r=\frac12\sqrt{D^2+E^2-4F}.$

The zero case is sometimes called a “point circle,” but it is not a circle with positive radius. In the negative case, the square root does not give a real radius.

3. Points and lines relative to a circle

Throughout this section, assume the circle has radius $r>0$.

A point and a circle

Calculate the distance $d$ from the point $P$ to the center:

• If $d>r$, the point lies outside the circle.
• If $d=r$, it lies on the circle.
• If $d<r$, it lies inside the circle.

A line and a circle

Write the line as $Ax+By+C=0$, where $A$ and $B$ are not both zero. The distance from the center $(a,b)$ to the line is

$d=\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}.$

• If $d<r$, there are two intersection points; the line is a secant.
• If $d=r$, there is one intersection point; the line is tangent to the circle.
• If $d>r$, there are no intersection points.

4. Tangent lines

The tangent at a point on the circle

If $P(x_0,y_0)$ lies on $x^2+y^2=r^2$, the tangent at $P$ is

$x_0x+y_0y=r^2.$

For a circle centered at $(a,b)$, the formula becomes

$(x_0-a)(x-a)+(y_0-b)(y-b)=r^2.$

Check that $P$ is on the circle before using the formula. The tangent is perpendicular to the radius at the point of contact.

Tangents through an external point

For an external point $(x_0,y_0)$, write a nonvertical candidate line as $y-y_0=k(x-x_0)$. Solve for $k$ by setting the distance from the center to the line equal to $r$.

Slope form does not include vertical lines, so also check $x=x_0$ separately. This line is tangent precisely when $|x_0-a|=r$.

5. Worked examples

Example 1: Use a center and a point on the circle

Find the equation of the circle centered at $(1,-2)$ and passing through $P(4,2)$.

The radius is

$r=\sqrt{(4-1)^2+(2-(-2))^2}=\sqrt{9+16}=5.$

The equation is therefore

$(x-1)^2+(y+2)^2=25.$

Example 2: Determine whether a line is tangent

Determine how $3x-4y+15=0$ meets the circle $x^2+y^2=9$.

The center is $(0,0)$ and the radius is $3$. The distance from the center to the line is

$d=\frac{|3(0)-4(0)+15|}{\sqrt{3^2+(-4)^2}}=\frac{15}{5}=3.$

Since $d=r$, the line is tangent to the circle.

Example 3: Find the center and radius from general form

Find the center and radius of $x^2+y^2-4x+6y-12=0$.

Using the formulas: With $D=-4$, $E=6$, and $F=-12$, the center is $(2,-3)$ and

$r=\frac12\sqrt{16+36-4(-12)}=\frac12\sqrt{100}=5.$

Completing the square:

$(x-2)^2-4+(y+3)^2-9-12=0,$

which gives

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

The center is $(2,-3)$ and the radius is $5$, agreeing with the first method.

Sources and review

Content reviewed: 12 September 2026.

• Paul's Online Notes: circle equations and completing the square
• Paul's Online Notes: why a negative sum of squares gives no real circle