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

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Answer 1 · 2017-04-15T16:40:18

See below.

This is a circle, and the standard form is:

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

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

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

How do you find the center and radius for x^2+(y-3)^2=16x2+(y−3)2=16x^2+(y-3)^2=16?