Algebra.Com's Answer #330970 by Edwin McCravy(20063)![\"\"](\"/images/stars/stars-13.gif\")  
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Find the equation of the circle. Touching the x-axis and passing through the points (3,1),(10,8).
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document.write( "We plot those points:\r\n" );
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document.write( "We can guess how to draw this circle using a compass. There are\r\n" );
document.write( "two possible circles:\r\n" );
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document.write( "Let the center of the circle be (h,k), then since the circle touches\r\n" );
document.write( "the x-axis, we draw a green vertical radius down to the x-axis:\r\n" );
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document.write( "Because it touches the x-axis, we can see from the graph above, k,\r\n" );
document.write( "which is the center's y-coordinate is the same as the radius r.\r\n" );
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document.write( "The equation of a crcle with center (h,k) and radius r has equation:\r\n" );
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document.write( "(x - h)² + (y - k)² = r²\r\n" );
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document.write( "And since k = r, it is\r\n" );
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document.write( "(x - h)² + (y - r)² = r²\r\n" );
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document.write( "x² - 2hx + h² + y² - 2ry + r² = r²\r\n" );
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document.write( "x² - 2hx + h² + y² - 2ry = 0 \r\n" );
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document.write( "We now substitute each of the given points in the equation:\r\n" );
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document.write( "Substituting (x,y) = (3,1)\r\n" );
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document.write( "3² - 2h(3) + h² + 1² - 2r(1) = 0\r\n" );
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document.write( "9 - 6h + h² + 1 - 2r = 0\r\n" );
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document.write( "10 - 6h + h² - 2r = 0\r\n" );
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document.write( "Substituting (x,y) = (10,8)\r\n" );
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document.write( "x² - 2hx + h² + y² - 2ry = 0 \r\n" );
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document.write( "10² - 2h(10) + h² + 8² - 2r(8) = 0\r\n" );
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document.write( "100 - 20h + h² + 64 - 16r = 0\r\n" );
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document.write( "164 - 20h + h² - 16r = 0\r\n" );
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document.write( "So we have this system of equations:\r\n" );
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document.write( " 10 - 6h + h² - 2r = 0\r\n" );
document.write( "164 - 20h + h² - 16r = 0\r\n" );
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document.write( "We eliminate r by multiplying the first equation by -8\r\n" );
document.write( "and adding the two equations:\r\n" );
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document.write( "-80 + 48h - 8h² + 16r = 0\r\n" );
document.write( "164 - 20h + h² - 16r = 0\r\n" );
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document.write( " 84 + 28h - 7h² - 0\r\n" );
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document.write( "Arrange in descending powers of h\r\n" );
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document.write( "-7h² + 28h + 84 = 0\r\n" );
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document.write( "Divide through by -7\r\n" );
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document.write( " h² - 4h - 12 = 0\r\n" );
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document.write( "Factor:\r\n" );
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document.write( "(h - 6)(h + 2) = 0\r\n" );
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document.write( "h - 6 = 0; h + 2 = 0\r\n" );
document.write( " h = 6; h = -2\r\n" );
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document.write( "To find the value of r corresponding to h = 6, substitute\r\n" );
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document.write( " 10 - 6h + h² - 2r = 0\r\n" );
document.write( "10 - 6(6) + 6² - 2r = 0 \r\n" );
document.write( " 10 - 36 + 36 - 2r = 0 \r\n" );
document.write( " 10 - 2r = 5 \r\n" );
document.write( " -2r = -10\r\n" );
document.write( " r = 5\r\n" );
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document.write( "So one possibility is the amaller circle \r\n" );
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document.write( "(x - h)² + (y - r)² = r²\r\n" );
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document.write( "which becomes\r\n" );
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document.write( "(x - 6)² + (y - 5)² = 5²\r\n" );
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document.write( "That's the standard form, which when multiplied out gives\r\n" );
document.write( "the general form:\r\n" );
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document.write( "x² + y² - 12x - 10y + 36 = 0\r\n" );
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document.write( "To find the value of r corresponding to h = -2, substitute\r\n" );
document.write( "h = 2 in\r\n" );
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document.write( " 10 - 6h + h² - 2r = 0\r\n" );
document.write( "10 - 6(-2) + (-2)² - 2r = 0 \r\n" );
document.write( " 10 + 12 + 4 - 2r = 0 \r\n" );
document.write( " 26 - 2r = \r\n" );
document.write( " -2r = -26\r\n" );
document.write( " r = 13\r\n" );
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document.write( "And the other possibility is the larger circle which is\r\n" );
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document.write( "(x - h)² + (y - r)² = r²\r\n" );
document.write( "(x + 2)² + (y - 13)² = 13²\r\n" );
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document.write( "That's the standard form, which when multiplied out gives\r\n" );
document.write( "the general form:\r\n" );
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document.write( "x² + y² + 4x - 25y + 4 = 0\r\n" );
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document.write( "Two possible solutions.\r\n" );
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document.write( "Edwin \n" );
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