How do you find the equation of the circle given Center at (0, 8); r = 8?

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Answer 1 · 2016-06-14T06:42:50

#x^2+y^2-16x=0#

Let the point on te circle be #(x,y)#.

As its distance from center #(0,8)# is always #8#, this being a circle with radius #8#, we have

#sqrt((x-0)^2+(y-8)^2)=8# or

#(x-0)^2+(y-8)^2=64# or

#x^2+y^2-16x+64=64# or

#x^2+y^2-16x=0#

graph{x^2+y^2-16x=0 [-12, 28, -10.08, 9.92]}