# Math Help: How to Find the Equation, Center, and Radius of a Circle

If you are reading this article, I’m assuming you know what a circle is, but for the sake of mathematical clarity, here’s the official definition:

“A **circle** is the set of all points in a plane equidistant from a fixed point. The fixed distances is called the **radius** , and the fixed point is called the **center** .”

In Geometry and Precalculus (and in some Algebra and Trigonometry classes) you are excepted to know the standard equation of a circle and use it to solve a couple common circle equation problems.

First off, here is the Standard Equation of a Circle:

A circle with radius *r* and center at *(0,0)* :

x2 + y2 = r2, where r > 0

A circle with radius *r* and center at *(h,k)* :

(x-h)2 + (y – k)2 = r^2, where r > 0

**Common Circle Equation Problem #1:**

Find the equation of a circle with radius 5 and center at:

a. (0,0)

b. (2, -1)

c. (-1,4)

This is fairly easy to answer, given the information above:

a. x2 + y2 = 25

b. (x-2)2 + (y+1)2 = 25

c. (x+1)2 + (y-4)2 = 25

**Common Circle Equation Problem #2:**

Find the center and radius of the circle with equation x2 + y2 + 6x – 4y = 23.

This is not as easy to answer, as it requires that we transform the given circle equation into the standard equation of a circle format. That is, we have to transform it into (x-h)2 + (y – k)2 = r2 by completing the square relative to *x* and relative to *y* .

This article will not go into great detail about the process of “completing the square”, though I will show the basic steps.

First group together the *x* and *y* parts and leave the constant on the right hand side of the equation.

(x2 + 6x + ) + (y2 – 4y + ) = 23

Notice we leave a spot for new values to be added in for the “complete the square” process.

(x2 + 6x + **9** ) + (y2 – 4y + **4** ) = 23 + **9** + **4**

To review, we got **9** by taking 6, dividing it by 2, and then squaring it. We got **4** by taking -4, dividing it by 2, then squaring it. To keep the equation balanced we had to add both new values on to the right hand side of the equation as well.

Each of the two subequations can now be rewritten as:

(x + 3)2 + (y-2)2 = 36

We now have our circle equation in standard format. From this we can easily pick out our *h* and *k* and *r* .

The circle is at point (-3, 2), with a radius of 6.

Blessings!

**Source **

*Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen.*Precalculus. Functions and Graphs. Fifth Edition

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