How do you find the center and radius of the circle given #x^2+y^2-3x+8y=20#?

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Answer 1 · 2016-11-24T08:39:41

The center is #(3/2,-4)#
The radius is #=6.185#

We have to rearrange the equation and complete the squares

#x^2+y^2-3x+8y=20#

#x^2-3x+y^2+8y=20#

#x^2-3x+9/4+y^2+8y+16=20+9/4+16#

#(x-3/2)^2+(y+4)^2=38.25=6.185^2#

So the center of the circle is #(3/2,-4)#
and the radius is #=6.185#

graph{x^2+y^2-3x+8y=20 [-14.28, 14.2, -10.85, 3.39]}