document.write( "Question 1088210: Find the equations of the lines described. Give your answers in the form y=mx+b. tangent lines to x²+y²-22x+12y+137=0 at points where x=15 \n" ); document.write( "
Algebra.Com's Answer #702459 by natolino_2017(77)\"\" \"About 
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first we need dy/dx, by using implicit derivative over the relation\r
\n" ); document.write( "\n" ); document.write( "2x + 2y(dy/dx) - 22 + 12dy/dx = 0\r
\n" ); document.write( "\n" ); document.write( "dy/dx = (11-x)/(y+6) for y different to -6.\r
\n" ); document.write( "\n" ); document.write( "then, we need to find the point(s) when x = 15 using the relation:\r
\n" ); document.write( "\n" ); document.write( "(15)^2 + y^2 - 22(15) + 12y + 137 = 0\r
\n" ); document.write( "\n" ); document.write( "Solving for y: y^2 + 12y + 32 = 0\r
\n" ); document.write( "\n" ); document.write( "has two solutions y = {-4, -8} So the point are P =(15,-4) and Q= (15,-8).\r
\n" ); document.write( "\n" ); document.write( "first when the point is P the dy/dx = (11-15)/(-4+6) = -2 (slope)\r
\n" ); document.write( "\n" ); document.write( "so the line is L : y = -2x +b, knowing that P belongs to the line.\r
\n" ); document.write( "\n" ); document.write( " -4 = -2(15) + b, so b = 26.\r
\n" ); document.write( "\n" ); document.write( "Second for the point Q, dy/dx = (11-15)/(-8+6) = 2 (slope)\r
\n" ); document.write( "\n" ); document.write( "so the line is L2: y = 2x + c, knowing that Q belongs to the line.\r
\n" ); document.write( "\n" ); document.write( "-8 = 2(15) +c, so c = -38.\r
\n" ); document.write( "\n" ); document.write( "Answer: L: y = -2x +26 , L2: y = 2x -38, which are tangent lines to the intersection of the curve with the line x = 15.
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