Given a circle $C(x,y) \equiv x^2 + y^2 + 2gx+2fy+c=0$ and a point $P = (x_1,y_1)$ outside the circle, the equation of the pair of straight lines that are tangent to the circle and pass through $P$ is given by $$C(x_1,y_1)C(x,y) = T(x,y)^2,$$
where $T(x,y)$ is the equation of the chord of contact of the tangents drawn from point $P$.
I do not know a derivation using Plucker's $\mu$. Please help, this question is driving me nuts. I am unable to understand the significance of squaring the equation of a line.
To find the equation of the pair of tangents, we have to find a second degree curve which passes through the intersection of $C=0$ and $T=0$ (say $A$ and $B$) and is tangent to $C$.
For this consider the curve $S(x,y)\equiv C(x,y)+\lambda T^2(x,y)$.
This curve will obviously always pass through $A$ and $B$ because at those points both $C=0$ and $T=0$ hold true. Now the reason we have squared the equation of the line is because we want the required curve to be tangent to $C$ at both $A$ and $B$. Have a look at the derivative of $S$: $$\frac{d}{dx}\left(S(x,y)\right)=\frac{d}{dx}\left(C(x,y)\right)+2\lambda T(x,y)\frac{d}{dx}\left(T(x,y)\right)$$ You can see that at the points $A$ and $B$ the derivative of $S$ is equal to the derivative of $C$ and hence $S$ and $C$ must be 'touching' at these points. Had we not squared the line's equation, this would not have been possible due to the absence of the extra $T(x,y)$.
After understanding the above, all you have to do is plug in the co-ordinates of $P$ in the curve $S$ since it passes through that point, by doing which we get:- $C(x_1,y_1)+\lambda T^2(x_1,y_1)=0$
You can check that for any point $P$, $C(x_1,y_1)=T(x_1,y_1)$ and this will result in $\lambda=-\dfrac 1{C(x_1,y_1)}$. Plugging the value of $\lambda$ completes the derivation.
EDIT: A couple of clarificatons :
For the purpose of this question the assumption is that we're using "standard" forms of the equations for the circle and the chord of contact, ie,
$$C\equiv x^2+y^2+2gx+2fy+c$$ and $$T\equiv xx_1+yy_1+g(x+x_1)+f(y+y_1)+c $$
Keeping that in mind, it's fairly obvious that $C(x_1,y_1)=T(x_1,y_1)$
Additionally, you can easily check that the discriminant for the resultant conic $CC_1 = T^2$, is zero if $P$ is the origin (it is a tedious task to show this for a general point, but it can be done), so I'm fairly confident that it is always a pair of straight lines.