# What are the coordinates of the center of the circle x^2 + y^2 -8x -10y -8=0?

May 17, 2016

$\left(4 , 5\right)$

#### Explanation:

Complete the squares for $x$ and $y$ as follows:

$0 = {x}^{2} + {y}^{2} - 8 x - 10 y - 8$

$= \textcolor{b l u e}{{x}^{2} - 8 x + 16} + \textcolor{g r e e n}{{y}^{2} - 10 y + 25} - 49$

$= \textcolor{b l u e}{{\left(x - 4\right)}^{2}} + \textcolor{g r e e n}{{\left(y - 5\right)}^{2}} - {7}^{2}$

Add ${7}^{2}$ to both ends and transpose to get:

${\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = {7}^{2}$

This is in the form of the standard equation of a circle:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

with center $\left(h , k\right) = \left(4 , 5\right)$ and radius $r = 7$.

graph{((x-4)^2+(y-5)^2-7^2)((x-4)^2+(y-5)^2-0.06)=0 [-16.67, 23.33, -5.2, 14.8]}