document.write( "Question 388037: Find the center of the circle using this equation: 4x^2+4y^2-24x-48y-16+0. \n" ); document.write( "

Algebra.Com's Answer #274354 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,         
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document.write( "Standard Form of an Equation of a Circle is 
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document.write( "where Pt(h,k) is the center and r is the radius
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document.write( "4x^2+4y^2-24x-48y-16 = 0
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document.write( "x^2+y^2-6x-12y-4 = 0
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document.write( "completing the squares to put into standard form
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document.write( " (x-3)^2 - 9 + (y-6)^2 - 36 - 4 = 0
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document.write( "  (x-3)^2 + (y-6)^2 = 49
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document.write( "Center Pt(3,6) radius = 7
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document.write( "CIRCLE
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