How to find the center and radius of #x^2 +(y+2)^2=72#?

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Answer 1 · 2018-08-05T00:34:53

Center @ #(0,-2)#, radius #6sqrt2#

Recall the equation of a circle

#bar( ul|color(white)(2/2)(x-h)^2+(y-k)^2=r^2color(white)(2/2)|)#, with center #(h,k)# and radius #r#.

With this in mind, we can rewrite our equation as

#(x-0)^2+(y--2)^2=sqrt72#

This tells us that our circle is centered at #(0,-2)#, and our radius is #sqrt72#, which can be simplified as

#sqrt72=sqrt(36*2)=sqrt36sqrt2=6sqrt2#

We are centered at #(0,-2)# and our radius is #6sqrt2#.

Hope this helps!