document.write( "Question 1141543: Find the center of the circles tangent to 5x-2y-1=0 at 1,2 with a radius of 3
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Algebra.Com's Answer #762115 by Edwin McCravy(20055)\"\" \"About 
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Let the center of the circle be (h,k). \r\n" );
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document.write( "The red line is the given line\r\n" );
document.write( "5x-2y-1 = 0.  The green line is the line perpendicular to it.\r\n" );
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document.write( "Since 5x-2y-1 = 0 solved for y is \"y=expr%285%2F2%29x-1%2F2\", its slope\r\n" );
document.write( "is \"5%2F2\" so the green line has slope \"-2%2F5\" and since it\r\n" );
document.write( "goes through (1,2) we use the point slope formula and get the equation:\r\n" );
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document.write( "\"y=y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
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document.write( "\"y-2=expr%28-2%2F5%29%28x-1%29\"\r\n" );
document.write( "\"5y-10=-2%28x-1%29\"\r\n" );
document.write( "\"5y-10=-2x%2B2\"\r\n" );
document.write( "\"2x%2B5y=12\" <-- equation of green line.\r\n" );
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document.write( "Since (h,k) lies on the green line,\r\n" );
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document.write( "\"2h%2B5k=12\"\r\n" );
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document.write( "Next we use the formula for the distance from a point (x1,y1) to the line\r\n" );
document.write( "Ax+By+C=0, which is\r\n" );
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document.write( "and the distance from (h,k) must be ±3 units to the line 5x-2y-1=0, so we have\r\n" );
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document.write( "\"matrix%281%2C2%2C%22%22+%2B-+abs%285h-2k-1%29%2Fsqrt%2829%29%2C%22%22%29=%22%22+%2B-+3\"\r\n" );
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document.write( "\"5h-2k-1+=+%22%22+%2B-+3sqrt%2829%29\"\r\n" );
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document.write( "\"5h-2k+=+1+%2B-+3sqrt%2829%29\"\r\n" );
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document.write( "We solve the system:\r\n" );
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document.write( "\"system%282h%2B5k=12%2C5h-2k+=+1+%2B-+3sqrt%2829%29%29\"\r\n" );
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document.write( "using the + and then the -, the two centers are\r\n" );
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document.write( "(h,k) = (-1.785439973,3.114172029)\r\n" );
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document.write( "and\r\n" );
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document.write( "(h,k) = (3.785430073,0.8858279709)\r\n" );
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document.write( "Edwin
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