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

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Answer 1 · 2015-02-14T15:10:55

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:

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

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