SOLUTION: Obtain the condition that lx+my+n =0 may be tangent to the circle x^2+y^2+2gx+2fy+c=0

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Question 1085694: Obtain the condition that lx+my+n =0 may be tangent to the circle x^2+y^2+2gx+2fy+c=0
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let the point of intersection be (u,v).
So then,
u%5E2%2Bv%5E2%2B2gu%2B2fv%2Bc=0
and
lu%2Bmv%2Bn=0
You know that the slope of the tangent line is equal to the value of the derivative at the intersection point.
Find the derivative using implicit differentiation,
2xdx%2B2ydy%2B2gdx%2B2fdy=0
%282x%2B2g%29dx%2B%282y%2B2f%29dy=0
%28y%2Bf%29dy=-%28x%2Bg%29dx
dy%2Fdx=-%28x%2Bg%29%2F%28y%2Bf%29
So then at (u,v),
m=dy%2Fdx=-%28u%2Bg%29%2F%28v%2Bf%29
Using the point slope form of a line,
y-v=-%28%28u%2Bg%29%2F%28v%2Bf%29%29%28x-u%29
%28y-v%29%28v%2Bf%29=-%28u%2Bg%29%28x-u%29
%28y-v%29%28v%2Bf%29%2B%28u%2Bg%29%28x-u%29=0
vy%2Bfy-v%5E2-fv%2Bux-u%5E2%2Bgx-gu=0
%28u%2Bg%29x%2B%28v%2Bf%29y-%28u%5E2%2Bv%5E2%2Bgu%2Bfv%29=0
Comparing,
l=u%2Bg
m=v%2Bf
n=-%28u%5E2%2Bv%5E2%2Bgu%2Bfv%29