Question

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

Answer 1

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(-a/2,-b/2)` and `r=sqrt35`.

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+ax+by+c=0`,

center:

`C(-a/2,-b/2)`,

radius:

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

In our case:

`x^2+y^2+6x-8y-11=0`,

so:

C(-3,4) and `r=sqrt(9+16+11)=sqrt35`.

The second way is completing the squares:

`x^2+6x+9-9+y^2-8y+16-16-11=0`

so:

`(x+3)^2+(y-4)^2=35`, and this is the equation of a circle given the center and the radius:

`(x-x_c)^2+(y-y_c)^2=r^2`.