How do you graph $x^2 + y^2 - 8x - 6y - 11 = 0$ ?

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Answer 1 · 2015-09-26T23:03:00

Refer to explanation

The equation $x^2 + y^2 - 8x - 6y - 11 = 0$ can be rewritten as follows

$(x^2-8x+16)+(y^2-6y+9)=11+16+9=> (x-4)^2+(y-3)^2=6^2$

which is a circle with center $(4,3)$ and radius $r=6$ hence

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How do you graph x^2 + y^2 - 8x - 6y - 11 = 0x^2 + y^2 - 8x - 6y - 11 = 0?