document.write( "Question 530469: Find the equation of the circle centered at (4,6) and passing through the point (2,2).\r
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Algebra.Com's Answer #350058 by jim_thompson5910(35256)\"\" \"About 
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\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\" Start with the general equation of a circle.\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-4%29%5E2%2B%28y-6%29%5E2=r%5E2\" Plug in \"h=4\" and \"k=6\" (since the center is the point (h,k) ).\r
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\n" ); document.write( "\n" ); document.write( "\"%282-4%29%5E2%2B%282-6%29%5E2=r%5E2\" Plug in \"x=2\" and \"y=2\" (this is the point that lies on the circle, which is in the form (x,y) ).\r
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\n" ); document.write( "\n" ); document.write( "\"%28-2%29%5E2%2B%28-4%29%5E2=r%5E2\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"4%2B16=r%5E2\" Square each term.\r
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\n" ); document.write( "\n" ); document.write( "\"20=r%5E2\" Add.\r
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\n" ); document.write( "\n" ); document.write( "So because \"h=4\", \"k=6\", and \"r%5E2=20\", this means that the equation of the circle with center (4,6) that goes through the point (2,2) is \r
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\n" ); document.write( "\n" ); document.write( "\"%28x-4%29%5E2%2B%28y-6%29%5E2=20\".
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