Ok, I'm sure all of you senior guys are tired of these problems, but this one is tricky. I'll stump you, I know it....if not for a few seconds:

The function

x^2 -4x +y^2 +3 = 0

find the equation of the 2 tangent lines in this function that pass through the origin.

My attempt:

As usual I would take the derivative

2x - 4 +2y dy/dx = 0

dy/dx= (2-x)/y

And NORMALLY I would plug in the coordinates of the other point (0,0) to get the slope "m" and put it into y=mx+b form. But division by 0 leads to some problems.

so I took the original equation and re-arranged for y, which got me:

y= sqrt[ 4x-x^2 -3]

and took the slope the old fashioned way

m=(y-0)/(x-0) which equals the derivative (2-x)/y

y/x = (2-x)/y

sqrt[ 4x-x^2 -3]/x = (2-x)/sqrt[ 4x-x^2 -3]

simplifying to:

4x-x^2 -3 = x (2-x)

x=3/2

but this is not the answer. The actual answer is +/- (sqrt3)/3

I know I should keep the x^2 because the answer is obviously a simplification after the quadratic formula has been used. Where did I go wrong?