# How do you find the center and radius for  x^2 + y^2 - 8x + 14y + 40 = 0?

Jun 30, 2016

circle centre (4, -7), radius 5

#### Explanation:

Simple answer: by using algebra and completing the square - TWICE, once for x and once for y.

We're starting with:
${x}^{2} + {y}^{2} - 8 x + 14 y + 40 = 0$

So we complete the squares. Will do separately so it's clearer, hopefully

For the x's
${x}^{2} - 8 x = {\left(x - 4\right)}^{2} - 16$

For the y's
${y}^{2} + 14 y = {\left(y + 7\right)}^{2} - 49$

pop these back into the original, and we get

${\left(x - 4\right)}^{2} - 16 + {\left(y + 7\right)}^{2} - 49 + 40 = 0$

${\left(x - 4\right)}^{2} + {\left(y + 7\right)}^{2} = 25 \textcolor{b l u e}{= {5}^{2}}$

that's a circle centre (4, -7), radius 5