How do you find the center and radius of the circle given #x^2+y^2+9x-8y+4=0#?

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Answer 1 · 2016-12-18T09:28:13

The center is #(-9/2,4)# and the radius #r=sqrt129/2#

We complete the squares and rearrange the equation

#x^2+9x+y^2-8y=-4#

#x^2+9x+81/4+y^2-8y+16=-4+81/4+16#

And now we factorise

#(x+9/2)^2+(y-4)^2=129/4#

We compare this equation to the standard equation of a circle

#(x-a)^2+(y-b)^2=r^2#

The centre is #(a,b)# and the radius #=r#

In our case,

The center is #(-9/2,4)# and the radius #=sqrt129/2#