How do you graph #x^2 + y^2 - 4x + 6y - 12 = 0#?

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Answer 1 · 2016-01-19T23:10:18

Rearrange into the standard form of the equation of a circle with centre #(2, -3)# and radius #5#.

#0 = x^2+y^2-4x+6y-12#

#=(x^2-4x+4)+(y^2+6y+9)-25#

#=(x-2)^2+(y+3)^2-5^2#

Add #5^2# to both ends and transpose to get:

#(x-2)^2+(y-(-3))^2 = 5^2#

This is in the form:

#(x-h)^2+(y-k)^2 = r^2#

the standard form of the equation of a circle with centre #(h, k) = (2, -3)# and radius #r=5#

graph{(x^2+y^2-4x+6y-12)((x-2)^2+(y+3)^2-0.02)=0 [-15.12, 16.92, -11.33, 4.69]}