# How do you find the center & radius of the circle with the given equation x²+y²-2x-10y=55 and then graph the equation?

Dec 22, 2015

center: $\left(1 , 5\right)$
radius: $9$

#### Explanation:

The general standard form for the equation of a circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{\mathmr{and} a n \ge}{a}\right)}^{2} + {\left(y - \textcolor{b r o w n}{b}\right)}^{2} = {\textcolor{c y a n}{r}}^{2}$ with center at $\left(\textcolor{\mathmr{and} a n \ge}{a} , \textcolor{b r o w n}{b}\right)$ and radius $\textcolor{c y a n}{r}$.

We will attempt to convert the given equation into this form.

Given:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} - 2 x - 10 y = 55$
Gather the $x$ and $y$ terms
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 2 x} + \textcolor{red}{{y}^{2} - 10 y} = \textcolor{g r e e n}{55}$
Complete the squares for the $x$ and $y$ sub-expressions.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 2 x + 1} + \textcolor{red}{{y}^{2} - 10 y + 25} = \textcolor{g r e e n}{55} + \textcolor{b l u e}{1} + \textcolor{red}{25}$
Re-write as squared binomials equal to a squared constant.
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{\mathmr{and} a n \ge}{1}\right)}^{2} + {\left(y - \textcolor{b r o w n}{5}\right)}^{2} = {\textcolor{c y a n}{9}}^{2}$

Comparing this to the general standard form, we see:
$\textcolor{w h i t e}{\text{XXX}}$the center is at $\left(\textcolor{\mathmr{and} a n \ge}{1} , \textcolor{b r o w n}{5}\right)$
and
$\textcolor{w h i t e}{\text{XXX}}$the radius is $\textcolor{c y a n}{9}$