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

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Answer 1 · 2016-10-20T06:37:35

#(3, 0)#

Radius#=4 #

Given -

#(x-3)^2+y^2=16#

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

#(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#

Then the center is -

#(3, 0)#

The radius is #=sqrta^2=sqrt16=4#

Look at the diagram