Geometry (B): READ: Equation of a Circle Reading

Equations and Graphs of Circles

A circle is defined as the set of all points that are the same distance from a single point called the center. This definition can be used to find an equation of a circle in the coordinate plane.

Let’s consider the circle shown below. As you can see, this circle has its center at point $(2, 2)$ and it has a radius of $\ 3$ .

All the points $x, y$ on the circle are a distance of $3\;\mathrm{units}$ away from the center of the circle.

We can express this information as an equation with the help of the Pythagorean Theorem. The right triangle shown in the figure has legs of length $x - 2$ and $y -2$ and hypotenuse of length $\ 3$ . We write:

$(x-2)^2 + (y-2)^2 = 9$

We can generalize this equation for a circle with center at point $(x_0, y_0)$ and radius $r$ .

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

Example 3

Find the center and radius of the following circles:

A. $(x-4)^2 + (y-1)^2 = 25$

B. $(x+1)^2 + (y-2)^2 = 4$

A. We rewrite the equation as: $(x-4)^2 + (y-1)^2 = 5^2.$ The center of the circle is at point $\ (4, 1)$ and the radius is $\ 5$ .

B. We rewrite the equation as: $(x - (-1))^2 + (y - 2)^2 = 2^2.$ The center of the circle is at point $\ (-1, 2)$ and the radius is $\ 2$ .

Example 4

Graph the following circles:

A. $x^2 + y^2 = 9$

B. $(x + 2)^2 + y^2 = 1$

In order to graph a circle, we first graph the center point and then draw points that are the length of the radius away from the center.

A. We rewrite the equation as: $(x-0)^2 + (y-0)^2=3^2.$ The center of the circle is point at $\ (0, 0)$ and the radius is $\ 3$ .

B. We rewrite the equation as: $(x-(-2))^2 + (y-0)^2 = 1^2.$ The center of the circle is point at $\ (-2, 0)$ and the radius is $\ 1$ .

Example 5

Write the equation of the circle in the graph.

From the graph we can see that the center of the circle is at point $\ (-2, 2)$ and the radius is $3\;\mathrm{units}$ long.

Thus the equation is:

$(x + 2)^2 + (y - 2)^2 = 9$

Example 6

Determine if the point $\ (1, 3)$ is on the circle given by the equation: $(x - 1)^2 + (y + 1)^2 = 16.$

In order to find the answer, we simply plug the point $(1, 3)$ into the equation of the circle.

$(1 - 1)^2 + (3 + 1)^2 &= 16\\ 0^2 + 4^2 &= 16$

The point $\ (1, 3)$ satisfies the equation of the circle.

Example 7

Find the equation of the circle whose diameter extends from point $A = (-3, -2)$ to $B = ( 1, 4).$

The general equation of a circle is: $(x - x_0)^2 + (y - y_0)^2 = r^2$

In order to write the equation of the circle in this example, we need to find the center of the circle and the radius of the circle.

Let’s graph the two points on the coordinate plane.

We see that the center of the circle must be in the middle of the diameter.

In other words, the center point is midway between the two points $A$ and $B$ . To get from point $A$ to point $B$ , we must travel $4\;\mathrm{units}$ to the right and $6\;\mathrm{units}$ up. To get halfway from point $A$ to point $B$ , we must travel $2\;\mathrm{units}$ to the right and $3\;\mathrm{units}$ up. This means the center of the circle is at point $(-3 + 2, -2 + 3)$ or $(-1, 1)$ .

We find the length of the radius using the Pythagorean Theorem:

$r^2 = 2^2 + 3^2 \Rightarrow r^2 = 13 \Rightarrow r = \sqrt{13}$

Thus, the equation of the circle is: $(x + 1)^2 + (y - 1)^2 = 13.$

Completing the Square:

You saw that the equation of a circle with center at point $(x_0, y_0)$ and radius $r$ is given by:

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

This is called the standard form of the circle equation. The standard form is very useful because it tells us right away what the center and the radius of the circle is.

If the equation of the circle is not in standard form, we use the method of completing the square to rewrite the equation in the standard form.

Example 8

Find the center and radius of the following circle and sketch a graph of the circle.

$x^2 - 4x + y^2 - 6y + 9 = 0$

To find the center and radius of the circle we need to rewrite the equation in standard form. The standard equation has two perfect square factors one for the $x$ terms and one for the $y$ terms. We need to complete the square for the $x$ terms and the $y$ terms separately.

$x^2 - 4x + \underline{\;\;\;\;} + y^2 - 6y + \underline{\;\;\;\;} + 9 = \underline{\;\;\;\;} + \underline{\;\;\;\;}$

To complete the squares we need to find which constants allow us to factors each trinomial into a perfect square. To complete the square for the $x$ terms we need to add a constant of $4$ on both sides.

$x^2 - 4x + \underline{\;4\;} + y^2 - 6y + \underline{\;\;\;\;} + 9 = \underline{\;4\;} + \underline{\;\;\;\;}$

To complete the square for the $y$ terms we need to add a constant of $9$ on both sides.

$x^2 - 4x + \underline{\;4\;} + y^2 - 6y + \underline{\;9\;} + 9 = \underline{\;4\;} + \underline{\;9\;}$

We can factor the separate trinomials and obtain:

$(x - 2)^2 + (y - 3)^2 + 9 = 13$

This simplifies as:

$(x - 2)^2 + (y - 3)^2 = 4$

You can see now that the center of the circle is at point $(2, 3)$ and the radius is $2$ .

The following questions are for your own review. The answers are listed below for you to check your work and understanding.

Review Questions:

1. $x^2 + y^2 = 1$

2. $(x - 3)^2 + (y + 5)^2 = 81$

$x^2 + (y - 2)^2 = 4$
$(x + 6)^2 + (y + 1)^2 =25$
Check that the point $\ (3, 4)$ is on the circle given by the equation $x^2 + (y - 4)^2 = 9.$
Check that the point $\ (-5, 5)$ is on the circle given by the equation $(x + 3)^2 + (y + 2)^2 = 50.$
Write the equation of the circle with center at $\ (2, 0)$ and radius $\ 4$ .
Write the equation of the circle with center at $\ (4, 5)$ and radius $\ 9$ .
Write the equation of the circle with center at $\ (-1, -5)$ and radius $\ 10$ .

For 14 and 15, write the equation of the circles.

In a circle with center $\ (4, 1)$ one endpoint of a diameter is $\ (-1, 3)$ . Find the other endpoint of the diameter.
The endpoints of the diameter of a circle are given by the points $A = (1, 4)$ and $B = (7, 2)$ . Find the equation of the circle.
A circle has center $(0, 4)$ and contains point $(1, 1)$ . Find the equation of the circle.
A circle has center $(-2, -2)$ and contains point $(4, 4)$ . Find the equation of the circle.
Find the center and the radius of the following circle: $x^2 + 8x + y^2 - 2y - 19 = 0$ .
Find the center and the radius of the following circle: $x^2 - 10x + y^2 + 6y -15 =0$ .
Find the center and the radius of the following circle: $x^2 + 20x + y^2 - 30y + 181 = 0$ .