What is the equation of the circle with a center at $(-8 ,3 )$ and a radius of $4$ ?

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Answer 1 · 2016-01-26T22:50:51

$(x+8)^2+(y-3)^2=16$

The standard form of a circle with a center at $(h,k)$ and a radius $r$ is

$(x-h)^2+(y-k)^2=r^2$

Since the center is $(-8,3)$ and the radius is $4$ , we know that

${(h=-8),(k=3),(r=4):}$

Thus, the equation of the circle is

$(x-(-8))^2+(y-3)^2=4^2$

This simplifies to be

$(x+8)^2+(y-3)^2=16$

What is the equation of the circle with a center at (-8 ,3 )(-8 ,3 ) and a radius of 4 4 ?