document.write( "Question 30361: Question:
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document.write( "Find the center and radius of x^2+y^2-6x-2y-6=0\r
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document.write( "Possible Answers:
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document.write( "(A) none of these
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document.write( "(B) center (3,1) radius 16
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document.write( "(C) center (3,1) radius 4
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document.write( "(D) center (-3,-1) radius 16\r
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document.write( "Thanks! \n" );
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Algebra.Com's Answer #17075 by Fermat(136)![]() ![]() You can put this solution on YOUR website! x^2+y^2-6x-2y-6=0 \r \n" ); document.write( "\n" ); document.write( "A circle, of radius r, and centred on the point (h,k) can be written as, \n" ); document.write( "(x-h)² + (y-k)² = r² \n" ); document.write( "expanding this expression, \n" ); document.write( "x² - 2hx + h² + y² - 2ky + k² = r² \n" ); document.write( "x² + y² - 2hx - 2ky + (h² + k² - r²) = 0 \n" ); document.write( "comparing this with the expression we were given, \n" ); document.write( "x^2+y^2-6x-2y-6=0 \n" ); document.write( "and comparing coefficients, \n" ); document.write( "-2h = -6 \n" ); document.write( "-2k = -2 \n" ); document.write( "h² + k² - r² = -6 \n" ); document.write( "From the 1st two eqns, \n" ); document.write( "h = 3 \n" ); document.write( "===== \n" ); document.write( "k = 1 \n" ); document.write( "===== \n" ); document.write( "substituting for h=3 and k=1 into the 3rd eqn, \n" ); document.write( "9 + 1 - r² = -6 \n" ); document.write( "-r² = -16 \n" ); document.write( "r² = 16 \n" ); document.write( "r = 4 \n" ); document.write( "===== \n" ); document.write( "We therefore have: (h,k) = (3,1) and r = 4 \n" ); document.write( "============================================ \n" ); document.write( "Ans:C \n" ); document.write( "===== \n" ); document.write( " \n" ); document.write( " |