How do you find the center and radius of the circle with the given equation x^2 + y^2 +4x - 8y - 44 = 0?

Aug 2, 2016

The Centre is $\left(- 2 , 4\right)$ and radius$= 8$.

Explanation:

Let us rewrite the given eqn. of circle as,

${x}^{2} + 4 x + {y}^{2} - 8 y = 44$, and, complete the squares on L.H.S..

$\therefore {x}^{2} + 4 x + 4 + {y}^{2} - 8 y + 16 = 44 + 4 + 16 = 64$.

$\therefore {\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = {8}^{2}$.

We compare this eqn. with $: {\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$, representing the

circle with centre $\left(h , k\right)$ and radius $r$.

Therefore, the Centre is $\left(- 2 , 4\right)$ and radius$= 8$.