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#.