document.write( "Question 1067878: Find the equation of a circle that has radius length {sqrt(30)} and is tangent to the line 3x+y-5=0 at the point (-1,8). \n" ); document.write( "
Algebra.Com's Answer #683075 by Edwin McCravy(20055) $\"\"$ $\"About$
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document.write( "There will be two solutions because the circle could be on either \r\n" );
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document.write( "The equation of a circle is:\r\n" );
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document.write( "(x-h)² + (y-k)² = r² with center (h,k) and radius r.\r\n" );
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document.write( "The radius is the distance from (h,k) to (-1,8).\r\n" );
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document.write( "We use the distance formula:\r\n" );
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document.write( "Square both sides:\r\n" );
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document.write( "The radius  is also the perpendicular distance from\r\n" );
document.write( "(h,k) to the line 3x+y-5=0\r\n" );
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document.write( "We use the formula:\r\n" );
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document.write( "The perpendicular distance from the point (x₁,y₁)\r\n" );
document.write( "to the line Ax+By+C=0 is d = \r\n" );
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document.write( " = \r\n" );
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document.write( " = \r\n" );
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document.write( " = \r\n" );
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document.write( " = \r\n" );
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document.write( " = \r\n" );
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document.write( "Substitute in\r\n" );
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document.write( "There is a lot of work here!  I'm not going to type the\r\n" );
document.write( "steps.  But you square those out and use the quadratic\r\n" );
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document.write( "Use the + and solve for h and get \r\n" );
document.write( "then\r\n" );
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document.write( "Then substitute that in:\r\n" );
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document.write( "So the center of one circle is \r\n" );
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document.write( "(h,k) = \r\n" );
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document.write( "To find the equation we substitute for h,k, and r in\r\n" );
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document.write( "Using the - sign, and doing the same thing:\r\n" );
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document.write( "(h,k) = \r\n" );
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document.write( "You can get the equation of that circle the same way.\r\n" );
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