document.write( "Question 243003: The equation of the tangent line to the curve x^2 + y^2 =169 at the point (5,-12) is
\n" ); document.write( "(A) 5y-12x= -120
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\n" ); document.write( "(C) 5x-12y= 169
\n" ); document.write( "(D) 12x+5y= 0
\n" ); document.write( "(E) 12x+5y= 169
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Algebra.Com's Answer #178018 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
First derive both sides of with respect to 'x' to get\r
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\n" ); document.write( "\n" ); document.write( "Now solve for y':\r
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\n" ); document.write( "\n" ); document.write( "So the slope is simply the negative quotient of the two coordinates. For the point (5,-12), the slope at that point is \"m=-5%2F%28-12%29=5%2F12\" (since \"x=5\" and \"y=-12\")\r
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\n" ); document.write( "\n" ); document.write( "Now recall that the point slope formula is \r
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\n" ); document.write( "\n" ); document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\" where 'm' is the slope and is the point in which the line goes through.\r
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\n" ); document.write( "\n" ); document.write( "\"y--12=%285%2F12%29%28x-5%29\" Plug in \"m=5%2F12\", \"x%5B1%5D=5\", and \"y%5B1%5D=-12\"\r
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\n" ); document.write( "\n" ); document.write( "\"y%2B12=%285%2F12%29%28x-5%29\" Rewrite \"y--12\" as \"y%2B12\"\r
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\n" ); document.write( "\n" ); document.write( "\"y%2B12=%285%2F12%29x%2B%285%2F12%29%28-5%29\" Distribute \"5%2F12\"\r
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\n" ); document.write( "\n" ); document.write( "\"y%2B12=%285%2F12%29x-25%2F12\" Multiply \"5%2F12\" and \"-5\" to get \"-25%2F12\"\r
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\n" ); document.write( "\n" ); document.write( "\"y=%285%2F12%29x-25%2F12-12\" Subtract 12 from both sides to isolate y\r
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\n" ); document.write( "\n" ); document.write( "\"y=%285%2F12%29x-169%2F12\" Combine like terms \"-25%2F12\" and \"-12\" to get \"-169%2F12\" \r
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\n" ); document.write( "\n" ); document.write( "\"12y=5x-169\" Multiply EVERY term by the LCD 12 to clear out the fractions.\r
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\n" ); document.write( "\n" ); document.write( "\"-5x%2B12y=-169\" Subtract 5x from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"5x-12y=169\" Multiply EVERY term by -1 to make the 'x' coefficient positive.\r
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\n" ); document.write( "\n" ); document.write( "So the equation of the tangent line is \"5x-12y=169\"
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