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 $=sqrt129/2$

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