What type of graph is produced by the equation \begin{align*}x^2-2x+y^2-4y+1=0\end{align*}? How can you turn this equation into graphing form in order to graph it?

### Equations of Circles

A **circle** is one example of a conic section. A circle is the set of all points that are equidistant from a center point. The general equation of a circle is \begin{align*}(x-h)^2+(y-k)^2=r^2 \end{align*} where \begin{align*}(h, k) \end{align*} is the center of the circle and \begin{align*}r\end{align*} is the radius of the circle.

You can derive the equation of a circle with the help of the Pythagorean Theorem.

#### Deriving the Equation of a Circle

1. Label the three sides of the right triangle below in terms of \begin{align*}x, y\end{align*}, and \begin{align*}r\end{align*}. Then, use the Pythagorean Theorem to write an equation that shows the relationship between the three sides.

The three sides of the triangle are \begin{align*}x\end{align*}, \begin{align*}y\end{align*}, and \begin{align*}r\end{align*}, where \begin{align*}r\end{align*} is the hypotenuse.

By the Pythagorean Theorem, \begin{align*}x^2+y^2=r^2\end{align*}.

2. Do all points on the circle from the previous problem satisfy the equation \begin{align*}x^2+y^2=r^2\end{align*}?

Yes, anywhere you move the point \begin{align*}(x, y)\end{align*}, the lengths of the sides of the triangle will still be \begin{align*}x\end{align*}, \begin{align*}y\end{align*}, and \begin{align*}r\end{align*}, so \begin{align*}x^2+y^2=r^2\end{align*} will still be true.

In fact, the set of points that make up the circle is exactly the set of points that satisfy the equation \begin{align*}x^2+y^2=r^2\end{align*}. This is why the general equation of a circle centered at the origin is \begin{align*}x^2+y^2=r^2\end{align*}, where \begin{align*}r\end{align*} is the radius of the circle.

3. Why does a circle with center \begin{align*}(h, k)\end{align*} and radius \begin{align*}r\end{align*} have the general equation \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}?

Pick a point on the circle and draw a right triangle connecting that point with the center of the circle.

The three sides of the triangle have lengths \begin{align*}x-h\end{align*}, \begin{align*}y-k\end{align*} and \begin{align*}r\end{align*}.

The relationship between the sides of the triangle is shown with the Pythagorean Theorem:

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

Points on the other side of the circle produce triangles with slightly different side lengths:

However, the relationship between the three sides of this triangle is the same, as shown below:

\begin{align*}(h-x)^2+(k-y)^2 &= r^2 \\ ((-1)(x-h) )^2+((-1)(y-k) )^2 &= r^2 \\ (-1)^2 (x-h)^2+(-1)^2 (y-k)^2 &= r^2 \\ (x-h)^2+(y-k)^2 &= r^2 \end{align*}

The set of points that satisfy the equation \begin{align*}(x-h)^2+(y-k)^2=r^2 \end{align*} are the set of points that make up the circle.

**Examples**

**Example 1**

Earlier, you were asked how you can turn \begin{align*}x^2-2x+y^2-4y+1=0\end{align*} into graphing form in order to graph it.

The equation \begin{align*}x^2-2x+y^2-4y+1=0 \end{align*} is the equation for a circle. You can tell it is a circle because there is both an \begin{align*}x^2 \end{align*} term and a \begin{align*}y^2 \end{align*} term, and the coefficients of each are the same. To turn this equation into the form \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}, you must complete the square twice. Remember that when completing the square, you are trying to figure out what number to add to make a perfect square trinomial. In order to maintain the equality of the equation, you must add the same numbers to both sides of the equation.

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

#### Example 2

What is the equation of a circle centered at \begin{align*}(-4, 5)\end{align*} with a radius of \begin{align*}\sqrt{2}\end{align*}?

\begin{align*}h=-4\end{align*}, \begin{align*}k=5\end{align*}, \begin{align*}r=\sqrt{2}\end{align*}. The equation of the circle is \begin{align*}(x-(-4) )^2+(y-5)^2= \left ( \sqrt{2} \right )^2\end{align*}. This can be simplified to \begin{align*}(x+4)^2+(y-5)^2=2\end{align*}. Notice that the signs of the 4 and the 5 in the equation are opposite from the signs of the 4 and the 5 in the center point.

#### Example 3

What are the center and radius of the circle described by the equation:

\begin{align*}x^2-4x+y^2+10y+13=0?\end{align*}

Complete the square to rewrite the equation in graphing form:

\begin{align*}x^2-4x+\underline{4}+y^2+10y+\underline{25} &=-13 \underline{+4+25} \\ (x-2)^2+(y+5)^2 &=16 \end{align*}

The center of the circle is \begin{align*}(2, -5)\end{align*} and the radius is 4.

#### Example 4

Graph the circle from #3.

### Review

Graph the following circles:

1. \begin{align*}(x+2)^2+(y-5)^2=1\end{align*}

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

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

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

5. \begin{align*}(x-2)^2+y^2=25\end{align*}

Find the center and radius of the circle described by each equation.

6. \begin{align*} x^2+4x+y^2+4y+7=0\end{align*}

7. \begin{align*}x^2-2x+y^2+12y+32=0\end{align*}

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

9. \begin{align*}x^2+8x+y^2+9=0\end{align*}

10. \begin{align*}x^2+2x+y^2-6y+1=0\end{align*}

11. Explain why a circle with radius \begin{align*}r\end{align*} that is centered at the origin will have the equation \begin{align*}x^2+y^2=r^2\end{align*}.

12. Use the Pythagorean Theorem to help explain why a circle with a center in the **second quadrant** with radius \begin{align*}r\end{align*} and center \begin{align*}(h, k)\end{align*} will have the equation \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}. *Note that in the second quadrant the value for* \begin{align*}h\end{align*} *will be negative and the value for* \begin{align*}k\end{align*} *will be positive.*

13. Use the Pythagorean Theorem to help explain why a circle with a center in the **third quadrant** with radius \begin{align*}r\end{align*} and center \begin{align*}(h, k) \end{align*} will have the equation \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}. *Note that in the third quadrant the value for \begin{align*}h\end{align*}* *will be negative and the value for* \begin{align*}k\end{align*} *will be negative.*

14. Write the equation of the following circle:

15. Write the equation of the following circle:

### Review (Answers)

To see the Review answers, open this PDF file and look for section 10.2.