SOLUTION: The equation {{{x^2-8x+y^2-6y-50=0}}} defines a circle with center (4,3) and radius ?

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Question 19584: The equation x%5E2-8x%2By%5E2-6y-50=0 defines a circle with center (4,3) and radius ?
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
The equation x%5E2-8x%2By%5E2-6y-50=0 defines a circle with center (4,3) and radius ?
x^2-8x+y^2-6y-50=0...
(x^2-8x)+(y^2-6y)-50=0
[x^2-2*4x+4^2-4^2]+[y^2-2*3y+3^2-3^2]-50=0
[(x-4)^2-16]+[(y-3)^2-9]-50=0
(x-4)^2+(y-3)^2=50+16+9=75=%28sqrt%2875%29%29%5E2
The standard for m of equation of a circle with centre at (h,k) and radius=r is given by
(x-h)^2+(y-k)^2=r^2...so comparing with the above we get centre of the circle as (4,3) and radius =%28sqrt%2875%29%29