How do you find the center and radius for $x^2+(y-3)^2=16$ ?

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Answer 1 · 2017-04-15T16:40:18

See below.

This is a circle, and the standard form is:

$(x-h)^2+(y-k)^2=r^2$ , where $(h,k)$ is the vertex, and $r$ is the radius.

The equation you want can be rewritten as:
$(x-0)^2+(y-3)^2=4^2$

So, the center is $(0,3)$ , and the radius is $4$ .

How do you find the center and radius for x^2+(y-3)^2=16x^2+(y-3)^2=16?