Question 358061
Find the equation of the tangent to the circle 
x^2+y^2-4y-1=0 
The the point (2,1)
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Find the derivative.
2x + 2yy' - 4y' = 0
y'(2y-4) = -2x
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y' = -2x/(2y-4)
This is the slope at every point (x,y)
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Let x = 2 ; Let y = 1 to find the slope at (2,1)
y' = -4/(-2) = 2
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Find the line with slope = 2 passing thru (2,1)
1 = -2(2)+b
b = 5
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Equation of the tangent at (2,1)
y = 2x+5
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Cheers,
Stan H.
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