SOLUTION: Factor to find the center and radius of the given equation:
a) x^2+y^2-8x=65
I understood how to solve this question and it came out to (x+4)^2+y^2=81
r=9 and (h.k)=(4,0)
My q
Algebra ->
Coordinate-system
-> SOLUTION: Factor to find the center and radius of the given equation:
a) x^2+y^2-8x=65
I understood how to solve this question and it came out to (x+4)^2+y^2=81
r=9 and (h.k)=(4,0)
My q
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Question 861325: Factor to find the center and radius of the given equation:
a) x^2+y^2-8x=65
I understood how to solve this question and it came out to (x+4)^2+y^2=81
r=9 and (h.k)=(4,0)
My question is how to solve this next one.
b) x^2+y^2-4x+14y=-17
I don't understand why there is a "14y" and what I am suppose to do with it.
I first began by subtracting 10 from "14y" and "-17" and it came out to
(x+2)^2+(y-2)^2=-27. I don't believe it is correct though because I cannot square -27 to get the radius. Should I add 50 to both sides instead to receive 47 as the radius? So confused... Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! x^2+y^2-4x+14y=-17
x^2-4x+y^2+14y=-17
complete the square for x and for y
x^2-4x+4+y^2+14y+49=-17+4+49
(x-2)^2+(y+7)^2=36
