**circle**is the collection of points that are the same distance from a single point. What is the connection between the Pythagorean Theorem and a circle?

### Graphing Circles

The single point that all the points on a circle are equidistant from is called the **center** of the circle. A circle does not have a focus or a directrix, instead it simply has a center. Circles can be recognized immediately from the general equation of a conic when the coefficients of \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*} are the same sign and the same value. Circles are not functions because they do not pass the vertical line test. The distance from the center of a circle to the edge of the circle is called the **radius** of the circle. The distance from one end of the circle through the center to the other end of the circle is called the **diameter**. The diameter is equal to twice the radius.

The graphing form of a circle is:

\begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}

The center** **of the circle is at \begin{align*}(h, k)\end{align*} and the radius of the circle is \begin{align*}r\end{align*}. Note that this looks remarkably like the Pythagorean Theorem.

To graph a circle, first plot the center and then apply the radius. Take the following equation for a circle:

\begin{align*}(x-1)^2+(y+2)^2=9\end{align*}

The center is at \begin{align*}\left(1, -2\right)\end{align*}. Plot that point and the four points that are exactly 3 units from the center.

### Examples

#### Example 1

Earlier, you were asked about the connection between circles and the Pythagorean Theorem. The reason why the graphing form of a circle looks like the Pythagorean Theorem is because each \begin{align*}x\end{align*} and \begin{align*}y\end{align*} coordinate along the outside of the circle forms a perfect right triangle with the radius as the hypotenuse.

#### Example 2

Graph the following conic: \begin{align*}(x+2)^2+(y-1)^2=1\end{align*}

#### Example 3

Turn the following equation into graphing form for a circle. Identify the center and the radius.

\begin{align*}36x^2+36y^2-24x+36y-275=0\end{align*}

Complete the square and then divide by the coefficient of \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*}

\begin{align*}& 36x^2-24x+36y^2+36y=275\\ & 36 \left(x^2-\frac{2}{3}x+\underline{\;\;\;\;}\right)+36 (y^2+y+\underline{\;\;\;\;})=275\\ & 36 \left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+36 \left(y^2+y+\frac{1}{4}\right)=275+4+9\\ & 36 \left(x-\frac{1}{3}\right)^2+36 \left(y+\frac{1}{2}\right)^2=288\\ & \left(x-\frac{1}{3}\right)^2+\left(y+\frac{1}{2}\right)^2=8\end{align*}

The center is \begin{align*}\left(\frac{1}{3}, -\frac{1}{2}\right)\end{align*}. The radius is \begin{align*}\sqrt{8}=2 \sqrt{2}\end{align*}.

#### Example 4

Write the equation for the following circle.

\begin{align*}(x-1)^2+(y+2)^2=4\end{align*}

**Example 5**

Write the equation of the following circle.

The center of the circle is at (3, 1) and the radius of the circle is \begin{align*}r=4\end{align*}. The equation is \begin{align*}(x-3)^2+(y-1)^2=16\end{align*}.

### Review

Graph the following conics:

1. \begin{align*}(x+4)^2+(y-3)^2=1\end{align*}

2. \begin{align*}(x-7)^2+(y+1)^2=4\end{align*}

3. \begin{align*}(y+2)^2+(x-1)^2=9\end{align*}

4. \begin{align*}x^2+(y-5)^2=8\end{align*}

5. \begin{align*}(x-2)^2+y^2=16\end{align*}

Translate the following conics from standard form to graphing form.

6. \begin{align*}x^2-4x+y^2+10y+18=0\end{align*}

7. \begin{align*}x^2+2x+y^2-8y+1=0\end{align*}

8. \begin{align*}x^2-6x+y^2-4y+12=0\end{align*}

9. \begin{align*}x^2+2x+y^2+14y+25=0\end{align*}

10. \begin{align*}x^2-2x+y^2-2y=0\end{align*}

Write the equations for the following circles.

11.

12.

13.

14.

15.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 9.3.