document.write( "Question 19584: The equation $\"x%5E2-8x%2By%5E2-6y-50=0\"$ defines a circle with center (4,3) and radius ? \n" ); document.write( "

Algebra.Com's Answer #9489 by venugopalramana(3286) $\"\"$ $\"About$

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The equation $\"x%5E2-8x%2By%5E2-6y-50=0\"$ defines a circle with center (4,3) and radius ?
\n" ); document.write( "x^2-8x+y^2-6y-50=0...
\n" ); document.write( "(x^2-8x)+(y^2-6y)-50=0
\n" ); document.write( "[x^2-2*4x+4^2-4^2]+[y^2-2*3y+3^2-3^2]-50=0
\n" ); document.write( "[(x-4)^2-16]+[(y-3)^2-9]-50=0
\n" ); document.write( "(x-4)^2+(y-3)^2=50+16+9=75= $\"%28sqrt%2875%29%29%5E2\"$
\n" ); document.write( "The standard for m of equation of a circle with centre at (h,k) and radius=r is given by
\n" ); document.write( "(x-h)^2+(y-k)^2=r^2...so comparing with the above we get centre of the circle as (4,3) and radius = $\"%28sqrt%2875%29%29\"$
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