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!

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