# How do you find the center and radius of the ellipse with standard equation x^2+6x+y^2-8y-11=0?

Feb 14, 2015

First of all this is not an equation of an ellipse, but it is the equation of a circle (even if a circle can be defined as a "particular" ellipse).

The answer is: $C \left(- \frac{a}{2} , - \frac{b}{2}\right)$ and $r = \sqrt{35}$.

It is possible to answer to the question in two ways.

The first, simplier, is remembering some formulas:

Equation of a circle:

${x}^{2} + {y}^{2} + a x + b y + c = 0$,

center:

$C \left(- \frac{a}{2} , - \frac{b}{2}\right)$,

r=sqrt((a/2)^2+(b/2)^2-c.

In our case:

${x}^{2} + {y}^{2} + 6 x - 8 y - 11 = 0$,

so:

C(-3,4) and $r = \sqrt{9 + 16 + 11} = \sqrt{35}$.

The second way is completing the squares:

${x}^{2} + 6 x + 9 - 9 + {y}^{2} - 8 y + 16 - 16 - 11 = 0$

so:

${\left(x + 3\right)}^{2} + {\left(y - 4\right)}^{2} = 35$, and this is the equation of a circle given the center and the radius:

${\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$.