document.write( "Question 1179955: Find the equation of the line tangent to the circle at the indicated point.
\n" ); document.write( "x^2+y^2-6x+8y-144 = 0 at point (8,8)
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Algebra.Com's Answer #809538 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-6x%2By%5E2%2B8y=144+\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x%5E2-6x%2Bb%5E2%29-b%5E2%2B%28y%5E2%2B8y%2Bb%5E2%29-b%5E2=144+\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x%5E2-6x%2B3%5E2%29-3%5E2%2B%28y%5E2%2B8y%2B4%5E2%29-4%5E2=144+\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2-9%2B%28y%2B4%29%5E2-16=144+\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2%2B%28y%2B4%29%5E2-25=144+\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2%2B%28y%2B4%29%5E2=144+%2B25\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2%2B%28y%2B4%29%5E2=169\"\r
\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2%2B%28y%2B4%29%5E2=13%5E2\"\r
\n" ); document.write( "\n" ); document.write( "centre is (\"3\",\"-4\") and the radius is \"r=13\"\r
\n" ); document.write( "\n" ); document.write( "slope of line through (\"3\",\"-4\") and (\"8\",\"8\") \r
\n" ); document.write( "\n" ); document.write( "\"m=%288-%28-4%29%29%2F%288-3%29\"\r
\n" ); document.write( "\n" ); document.write( "\"m=%288%2B4%29%2F5\"\r
\n" ); document.write( "\n" ); document.write( "\"m=12%2F5\"\r
\n" ); document.write( "\n" ); document.write( "since tangent is perpendicular to it, the slope of line of the tangent is therefore \"-1%2F%2812%2F5%29=-5%2F12\"\r
\n" ); document.write( "\n" ); document.write( "So line is \"y=-%285%2F12%29x%2Bb\"\r
\n" ); document.write( "\n" ); document.write( "So, tangecy point is at (\"8\",\"8\")\r
\n" ); document.write( "\n" ); document.write( "\"8=-%285%2F12%298%2Bb\"\r
\n" ); document.write( "\n" ); document.write( "\"b+=+34%2F3\"\r
\n" ); document.write( "\n" ); document.write( "\"y=-%285%2F12%29x%2B34%2F3\"=> your tangent\r
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