How do you graph x2+y2−8x+6y+16=0?

1 Answer
George C.
Nov 1, 2015

0=x2+y2−8x+6y+16=(x−4)2+(y+3)2−32

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

Explanation:

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

(x−a)2+(y−b)2=r2

We are given:

0=x2+y2−8x+6y+16

=x2−8x+16+y2+6y+9−9

=(x−4)2+(y+3)2−32

So:

(x−4)2+(y+3)2=32

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]}

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