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### Section 3.3 : Circles

7. Determine the radius and center of the following circle. If the equation is not the equation of a circle clearly explain why not.

$9{x^2} + 9{y^2} - 6x - 36y - 107 = 0$

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To do this problem we need to complete the square on the $$x$$ and $$y$$ terms. To help with this we’ll first get the number on the right side and group the $$x$$ and $$y$$ terms as follows.

$9{x^2} - 6x + 9{y^2} - 36y = 107$

Also, we know that we want the coefficients of the $${x^2}$$ and $${y^2}$$ to be one for the completing the square process so it will be just as easy to divide everything by a 9 to help out with the completing the square.

${x^2} - \frac{2}{3}x + {y^2} - 4y = \frac{{107}}{9}$ Show Step 2

Here are the numbers we need to complete the square for both $$x$$ and $$y$$.

$x:\,{\left( {\frac{{ - {}^{2}/{}_{3}}}{2}} \right)^2} = {\left( { - \frac{1}{3}} \right)^2} = \frac{1}{9}\hspace{0.55in}y:\,{\left( {\frac{{ - 4}}{2}} \right)^2} = {\left( { - 2} \right)^2} = 4$ Show Step 3

Now, complete the square.

\begin{align*}{x^2} - \frac{2}{3}x + \frac{1}{9} + {y^2} - 4y + 4 & = \frac{{107}}{9} + \frac{1}{9} + 4\\ {\left( {x - \frac{1}{3}} \right)^2} + {\left( {y - 2} \right)^2} & = 16\end{align*}

Don’t forget to add the numbers from Step 2 to both sides of the equation!

Show Step 4

So, we have the equation in standard form and so we can quickly identify the radius and center of the circle.

${\mbox{Radius : }}r = 4\hspace{0.25in}\hspace{0.25in}{\mbox{Center : }}\left( {\frac{1}{3},2} \right)$

If you don’t recall how to get the radius and center from the standard form of a circle check out Problems 3 – 5 in this section for some practice.