SOLUTION: the equation of tangent at a point (5,1) to the circle x^2+y^2-2x+4y-20=0 will be???

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Question 884621: the equation of tangent at a point (5,1) to the circle x^2+y^2-2x+4y-20=0 will be???
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Find the value of the derivative at the point.
The derivative is equal to the slope of the tangent line.
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x%5E2%2By%5E2-2x%2B4y-20=0
2xdx%2B2ydy-2dx%2B4dy=0
2ydy%2B4dy=2dx-2xdx
%282y%2B4%29dy=%282-2x%29dx
dy%2Fdx=%282-2x%29%2F%282y%2B4%29=%281-x%29%2F%28y%2B2%29
At (5,1),
m=dy%2Fdx=%281-5%29%2F%281%2B2%29
m=-4%2F3
Using the point slopt form of a line,
y-1=-%284%2F3%29%28x-5%29%7D%7D%0D%0A%7B%7B%7By=-%284%2F3%29x%2B20%2F3%2B3%2F3
highlight%28y=-%28-4%2F3%29x%2B23%2F3%29
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