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

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Answer 1 · 2016-10-20T06:37:35

$(3, 0)$ $(3, 0)$

Radius $= 4$ $=4$

Given -

${(x - 3)}^{2} + y^{2} = 16$ $(x-3)^2+y^2=16$

The equation of the circle with center $(h, k)$ $(h,k)$ is -

${(x - h)}^{2} + {(y - k)}^{2} = a^{2}$ $(x-h)^2+(y-k)^2=a^2$

In our case to find the center, rewrite the equation like this-

${(x - 3)}^{2} + {(y - 0)}^{2} = 16$ $(x-3)^2+(y-0)^2=16$

Then the center is -

$(3, 0)$ $(3, 0)$

The radius is $= {\sqrt{a}}^{2} = \sqrt{16} = 4$ $=sqrta^2=sqrt16=4$

Look at the diagram

How do you find the center and radius of the circle given (x−3)2+y2=16(x-3)^2+y^2=16?