# How do you find the center and radius of the circle x^2 + y^2 – 8x – 6y – 75 = 0?

Jul 4, 2016

Centre: $\left(4 , 3\right)$
Radius: $10$

#### Explanation:

Complete the square for $x$ and $y$ and rearrange:

$0 = {x}^{2} + {y}^{2} - 8 x - 6 y - 75$

$= \left({x}^{2} - 8 x + 16\right) + \left({y}^{2} - 6 y + 9\right) - 100$

$= {\left(x - 4\right)}^{2} + {\left(y - 3\right)}^{2} - {10}^{2}$

Add ${10}^{2}$ to both ends and transpose to get:

${\left(x - 4\right)}^{2} + {\left(y - 3\right)}^{2} = {10}^{2}$

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

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

where $\left(h , k\right) = \left(4 , 3\right)$ is the centre of the circle and $r = 10$ is the radius.