How do you graph #x^2+y^2-8x+6y+16=0#?
1 Answer
Nov 1, 2015
is a circle of radius
Explanation:
The equation of a circle of radius
#(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
graph{(x-4)^2+(y+3)^2 = 3^2 [-7.875, 12.125, -7.8, 2.2]}