Asked by Anonymous
determine the center of the circle whose equation is x^2+y^2+8x-12y+34?
Answers
Answered by
Steve
x^2+y^2+8x-12y+34 = 0
x^2+8x + y^2-12y = -34
(x^2+8x+16)+(y^2-12y+36) = -34+16+36
(x+4)^2 + (y-6)^2 = 18
...
x^2+8x + y^2-12y = -34
(x^2+8x+16)+(y^2-12y+36) = -34+16+36
(x+4)^2 + (y-6)^2 = 18
...
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