How do you graph $x^2+y^2+8x-8y-93=0$ ?

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Answer 1 · 2016-02-04T02:09:41

You can transfrom it as follows

$x^2+8x+16+y^2-8y+16-93-32=0=>(x+4)^2+(y-4)^2=125=> (x+4)^2+(y-4)^2=(5sqrt5)^2$

which is a circle with centre at $(-4,4)$ and radius $r=5sqrt5$

The graph is

enter image source here

How do you graph x^2+y^2+8x-8y-93=0x^2+y^2+8x-8y-93=0?