document.write( "Question 1093746: The center of a circle is located at O(1, h). The line whose equation is y=kx+1 is tangent to circle O at the point P(3,6). Find the values of k and h.\r
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Algebra.Com's Answer #708373 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Comparing with the slope intercept form y=mx+b, the tangent line IP\r\n" );
document.write( "whose equation is y=kx+1 has y-intercept (0,b) = I(0,1), and \r\n" );
document.write( "slope m=k.\r\n" );
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document.write( "Since the line goes through I(0,1) and P(3,6), we use the\r\n" );
document.write( "slope formula to determine m = k, its slope:\r\n" );
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document.write( "\"m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29\"\r\n" );
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document.write( "\"k=%286-1%29%2F%283-0%29=5%2F3\"  <-- that's the answer for k\r\n" );
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document.write( "The radius OP is perpendicular to line IP, so its slope is\r\n" );
document.write( "the reciprocal of \"5%2F3\" with the opposite sign, so the\r\n" );
document.write( "slope of the radius OP is \"-3%2F5\"\r\n" );
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document.write( "To find the equation of OP, we use the point-slope form of \r\n" );
document.write( "a line with \"m=-3%2F5\" and the point of tangency, P(3,6).\r\n" );
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document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
document.write( "\"y-6=expr%28-3%2F5%29%28x-3%29\"  <--equation of radius OP\r\n" );
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document.write( "Since center O(1,h) is on that radius, we substitute the\r\n" );
document.write( "point (1,h), to find h\r\n" );
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document.write( "\"h-6=expr%28-3%2F5%29%281-3%29\"\r\n" );
document.write( "\"h-6=expr%28-3%2F5%29%28-2%29\"\r\n" );
document.write( "\"h-6=6%2F5\"\r\n" );
document.write( "\"h=6%2B6%2F5\"\r\n" );
document.write( "\"h=30%2F5%2B6%2F5\"\r\n" );
document.write( "\"h=36%2F5\"     <---that's the answer for h\r\n" );
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document.write( "Edwin
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