@mertsj find the center and radius of the given circle. x^2+y^2-8x+12y-8=0

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@mertsj find the center and radius of the given circle. x^2+y^2-8x+12y-8=0

Mathematics
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aka complete the square (twice)
you can do this by converting your equation to the form (x - a)^2 + (y - b)^2 = r^2 where the center is (a,b) and radius = r you do this by a process called completing the square
ok so I have (x^2-8x)+(y^2+12y)=8

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ok now convert x^6 - 8x and y^2 + 12y to the form (x - a)^2 - c
and (y- b)^2 - d
(x^2-8x+16)+(y^2+12y+36)=8+16+36
x^2 - 8x = (x - 4)^2 - 16
and y^2 + 12y = (y + 6)^2 - 36
so we get (x -4)^2 + (y + 6)^2 = 8+16+36 = 60
gotcha
good
compare the above withe standard equation and you'll get what you want

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