document.write( "Question 908274: Find the sqaure of the length of the tangent to the point (-4,1) on the circle given by the equation x^2+y^2+8x+12y+5=0. \n" ); document.write( "
Algebra.Com's Answer #550966 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "First we put the circle's equation, which is in GENERAL form: \r\n" );
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document.write( "\"x%5E2%2By%5E2%2B8x%2B12y%2B5=0\"\r\n" );
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document.write( "in STANDARD form\r\n" );
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document.write( "\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\"\r\n" );
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document.write( "\"x%5E2%2By%5E2%2B8x%2B12y%2B5=0\"\r\n" );
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document.write( "Rearrange the terms so that the x term is just right of the x2 term,\r\n" );
document.write( "and the y term is just right of the y2 term, and the\r\n" );
document.write( "constant term is on t he right.\r\n" );
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document.write( "\"x%5E2%2B8x%2By%5E2%2B12y=-5\"\r\n" );
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document.write( "Now we do some work to the side to find the number necessary to \r\n" );
document.write( "complete the square on \"x%5E2%2B8x\":\r\n" );
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document.write( "1. Multiply the coefficient of x, which is 8, by \"1%2F2\", getting 4.\r\n" );
document.write( "2. Square 4, getting 16.\r\n" );
document.write( "3. Add +16 just after 8x on the left and also add +16 to the right side.\r\n" );
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document.write( "\"x%5E2%2B8x%2B16%2By%5E2%2B12y=-5%2B16\"\r\n" );
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document.write( "Also to the side we find the number necessary to \r\n" );
document.write( "complete the square on \"y%5E2%2B12y\":\r\n" );
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document.write( "1. Multiply the coefficient of y, which is 12, by \"1%2F2\", getting 6.\r\n" );
document.write( "2. Square 6, getting 36.\r\n" );
document.write( "3. Add +36 just after 12y on the left and also add +36 to the right side.\r\n" );
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document.write( "\"x%5E2%2B8x%2B16%2By%5E2%2B12y%2B36=-5%2B16%2B36\"\r\n" );
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document.write( "Factor the first three terms on the left as \"%28x%2B4%29%28x%2B4%29\" and write as\r\n" );
document.write( "\"%28x%2B4%29%5E2\"\r\n" );
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document.write( "Factor the last three terms on the left as \"%28y%2B6%29%28y%2B6%29\" and write as\r\n" );
document.write( "\"%28y%2B6%29%5E2\" \r\n" );
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document.write( "Combine the numbers on the right as 35.\r\n" );
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document.write( "Now we have the standard form of the circle:\r\n" );
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document.write( "\"%28x%2B4%29%5E2%2B%28y%2B6%29%5E2=47\"\r\n" );
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document.write( "Comparing to\r\n" );
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document.write( "\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\"\r\n" );
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document.write( "We see that -h=+4 or h=-4, -k=+6 or k=-6, r2=47,\r\n" );
document.write( "r=\"sqrt%2847%29\".\r\n" );
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document.write( "Now we can sketch the graph of the circle with center O(-4,-6) \r\n" );
document.write( "and radius \"sqrt%2847%29\". Wow, as we see, the point P(-4,1) is\r\n" );
document.write( "very close to the circle!. I'll draw the tangent line PT in red,\r\n" );
document.write( "and the radius to the point of tangency and the line from the \r\n" );
document.write( "center to P(4,1) in green.  We want to find the length of PT.\r\n" );
document.write( "The line OP is 7 units because it's directly above O, which is 6\r\n" );
document.write( "units below the x-axis, to P which is 1 unit above the x-axis, so\r\n" );
document.write( "OP is 6+1 or 7 units, and OT is \"sqrt%2847%29\". Triangle POT is a\r\n" );
document.write( "right triangle, so by the Pythagorean theorem:\r\n" );
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document.write( "\"OT%5E2%2BPT%5E2=OP%5E2\"\r\n" );
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document.write( "\"%28sqrt%2847%29%29%5E2%2BPT%5E2=7%5E2\"\r\n" );
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document.write( "\"47%2BPT%5E2=49\"\r\n" );
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document.write( "\"PT%5E2=2\"\r\n" );
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document.write( "\"PT=sqrt%282%29\"\r\n" );
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document.write( "Edwin
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