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Find the radius and coordinates of the center of the circle x^2+y^2+(5/6)x-(8/5)y-8=0
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Start by grouping the x terms together, then complete the square to get some (x-h)^2 term. Then do the same with the y terms to get (y-k)^2. Your center is at (h,k), and the radius will be the square root of the constant term (that you will have moved to the right to get the standard form for a circle (x-h)^2+(y-k)^2=r^2)
coordinates are -g = -5/12, -f=4/5. and radius = sqroot(g^2+f^2-c). g .f , and c can be find by comparing the equation by slandered equation of second order.
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