Equation of Circle

Locating a Circle from Its Center and Radius

A circle equation describes every point on a circle in the coordinate plane. Imagine drawing with a compass on graph paper: the needle fixes the center, the opening fixes the radius, and the pencil traces exactly the points that satisfy the equation.

The circle equation contains the coordinates of the center and the radius, so both can be determined without drawing the circle first.

Circle Centered at Origin

Start with the simplest case, a circle centered at the origin $O(0,0)$.

If we have a circle with center at $O(0,0)$ and radius $r$, then every point $(x,y)$ on that circle has the same distance from the center, which is $r$.

Using the distance formula, we get:

If we square both sides, we get the circle equation with center at origin:

The graph shows that every point has the same distance from the center.

For the circle above, the equation is $x^2 + y^2 = 9$ because the radius is $3$, so $r^2 = 3^2 = 9$.

Circle with Arbitrary Center

For a circle centered at $P(a,b)$, let the radius be $r$.

Every point $(x,y)$ on this circle must have the same distance $r$ from center $P(a,b)$. Using the distance formula:

After squaring, we get the standard equation for a circle with an arbitrary center:

The shifted center is visible in the graph below.

The equation of the circle above is $(x-2)^2 + (y+1)^2 = 4$ because the center is $P(2,-1)$ and the radius is $2$, so $r^2 = 2^2 = 4$.

General Form of Circle Equation

Sometimes we find circle equations that have already been expanded into general form. For example, from the equation $(x-1)^2 + (y-2)^2 = 9$, if we expand it:

This last form is called the general form of circle equation:

An equation in general form can be converted back to standard form by completing the square.

Not all equations of the form $x^2 + y^2 + Dx + Ey + F = 0$ are circle equations. The condition is $D^2 + E^2 - 4F > 0$. If this value is zero, then it's just a single point, and if negative, then there's no curve at all.

Completing the square shows where the center and radius formulas come from:

Determining Center and Radius

From the standard form $(x-a)^2 + (y-b)^2 = r^2$, we can directly know:

The center is at $(a,b)$ and the radius is $r$ (obtained from $r = \sqrt{r^2}$).

While from the general form $x^2 + y^2 + Dx + Ey + F = 0$, we can determine:

Practice Problems

Each problem gives the centre and radius, three points, or a general equation, and asks for the circle equation, so choose the form that matches the given information.

1. Determine the equation of a circle centered at $(3,-2)$ with radius $5$.

2. Given a circle with equation $x^2 + y^2 - 6x + 4y - 12 = 0$. Determine the center and radius of the circle.

3. A circle passes through point $(6,-8)$ and is centered at $O(0,0)$. Determine the equation of the circle.

4. Determine the equation of a circle that has a diameter with endpoints at $A(1,3)$ and $B(7,-1)$.

Worked Solutions

1. Solution:

   Given center $(a,b) = (3,-2)$ and radius $r = 5$.

   Using the circle equation formula:

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

   $$(x-3)^2 + (y-(-2))^2 = 5^2$$

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

   So the circle equation is $(x-3)^2 + (y+2)^2 = 25$.

2. Solution:

   From the equation $x^2 + y^2 - 6x + 4y - 12 = 0$, we identify the coefficients:

   $D = -6$, $E = 4$, $F = -12$

   $$\begin{aligned} &\text{Center: } \left(-\frac{D}{2}, -\frac{E}{2}\right) \\ &= \left(-\frac{(-6)}{2}, -\frac{4}{2}\right) \\ &= (3, -2) \end{aligned}$$

   $$\begin{aligned} r &= \frac{1}{2}\sqrt{D^2 + E^2 - 4F} \\ &= \frac{1}{2}\sqrt{(-6)^2 + 4^2 - 4(-12)} \end{aligned}$$

   $$\begin{aligned} r &= \frac{1}{2}\sqrt{36 + 16 + 48} \\ &= \frac{1}{2}\sqrt{100} \\ &= \frac{1}{2} \times 10 \\ &= 5 \end{aligned}$$

   So the circle center is at $(3,-2)$ with radius $5$.

3. Solution:

   Since the circle is centered at $O(0,0)$ and passes through point $(6,-8)$, the radius is the distance from the center to that point.

   $$\begin{aligned} r &= \sqrt{(6-0)^2 + (-8-0)^2} \\ &= \sqrt{6^2 + (-8)^2} \\ &= \sqrt{36 + 64} \\ &= \sqrt{100} \\ &= 10 \end{aligned}$$

   The circle equation: $x^2 + y^2 = r^2 = 10^2 = 100$

4. Solution:

   The circle center is the midpoint of diameter $AB$:

   $$\begin{aligned} &\text{Center} \\ &= \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) \\ &= \left(\frac{1+7}{2}, \frac{3+(-1)}{2}\right) \\ &= \left(\frac{8}{2}, \frac{2}{2}\right) \\ &= (4, 1) \end{aligned}$$

   The radius is half the diameter length:

   $$\begin{aligned} &\text{Length } AB \\ &= \sqrt{(x_B-x_A)^2 + (y_B-y_A)^2} \\ &= \sqrt{(7-1)^2 + (-1-3)^2} \end{aligned}$$

   $$\begin{aligned} &= \sqrt{6^2 + (-4)^2} \\ &= \sqrt{36 + 16} \\ &= \sqrt{52} \\ &= \sqrt{4 \times 13} \\ &= 2\sqrt{13} \end{aligned}$$

   $$r = \frac{\text{diameter}}{2} = \frac{2\sqrt{13}}{2} = \sqrt{13}$$

   The circle equation: $(x-4)^2 + (y-1)^2 = (\sqrt{13})^2 = 13$