How do you graph $x^2+y^2-8x+6y+16=0$ ?

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Answer 1 · 2015-11-01T10:38:13

$0 = x^2+y^2-8x+6y+16 = (x-4)^2 + (y+3)^2 - 3^2$

is a circle of radius $3$ with centre $(4, -3)$

The equation of a circle of radius $r$ centred at $(a, b)$ can be written:

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

We are given:

$0 = x^2+y^2-8x+6y+16$

$=x^2-8x+16 + y^2+6y+9 - 9$

$=(x-4)^2 + (y+3)^2 - 3^2$

So:

$(x-4)^2+(y+3)^2 = 3^2$

which is in the form of the equation of a circle of radius $3$ centre $(4, -3)$

graph{(x-4)^2+(y+3)^2 = 3^2 [-7.875, 12.125, -7.8, 2.2]}

How do you graph x^2+y^2-8x+6y+16=0x^2+y^2-8x+6y+16=0?