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]}