document.write( "Question 770881: Find the equation of the tangent to the circle with centre: (0,2) at (-1,5) \n" ); document.write( "
Algebra.Com's Answer #469826 by stanbon(75887)\"\" \"About 
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Find the equation of the tangent to the circle with centre: (0,2) at (-1,5)
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\n" ); document.write( "radius = sqrt[(5-2)^2+(-1-0)^2] = sqrt[10]
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\n" ); document.write( "Equation of the circle:
\n" ); document.write( "x^2 + (y-2)^2 = 10
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\n" ); document.write( "Solve for \"y\":
\n" ); document.write( "y^2-4y+4 + x^2 = 10
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\n" ); document.write( "Find dy/dx
\n" ); document.write( "2y(dydx) -4 + 2x = 0
\n" ); document.write( "dydx = (-2x+4)/(2y)
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\n" ); document.write( "dy/dx at (-1,5) is :: (2+4)/-2 = -3
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\n" ); document.write( "Equation of the line with slope = -3 passing thru (-1,5)
\n" ); document.write( "y = mx + b
\n" ); document.write( "Solve for \"b\":
\n" ); document.write( "5 = -3*-1 + b
\n" ); document.write( "b = 2
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\n" ); document.write( "Equation:
\n" ); document.write( "y = -3x + 2
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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