Direct/transverse tangents of two circles If equations of two circles in a plane are given as
$$ f(x,y)=0,\, g(x,y)=0,\, $$


*

*Find equations of direct tangents in terms of $f,g$


and 


*Find equations of transverse tangents in terms of $f,g$


We could calculate using coincident roots/point etc. of a transverse line but how can it be found  using pole/polar properties ? or other methods?
 A: If $f(x,y)=0$ is the equation of a circle with center $(h,k)$ and radius $r,$ then the most general form of $f$ is
$$f(x,y) = a((x-h)^2 + (y-k)^2 - r^2)$$
where $a\neq0.$ Consider the following values of $f$:
\begin{align}
f(0,0)  &= a(h^2 + k^2 + r^2), \\
f(1,0)  &= a(h^2 - 2h + 1 + k^2 + r^2), \\
f(-1,0) &= a(h^2 + 2h + 1 + k^2 + r^2).
\end{align}
We can combine these as follows:
$$
f(1,0) + f(-1,0) - 2f(0,0) = 2a,
$$
from which we find $a$ in terms of $f,$ namely,
$a = \tfrac12(f(1,0) + f(-1,0)) - f(0,0).$
We also have
$$
f(-1,0) - f(1,0) = 4ah,
$$
and since we can already express $a$ in terms of $f,$ this lets us express $h$ in terms of $f.$
We can also plug the expression for $a$ into the equation
$f(0,0) - f(0,1) = a(2k - 1)$ in order to find $k$ in terms of $f,$
and we can then plug these expressions for $a,$ $h,$ and $k$
into the equation $f(0,0) = a(h^2 + k^2 + r^2)$ to find $r$ in terms of $f.$
Since this establishes that $a,$ $h,$ $k,$ and $r$ can all be expressed in terms of $f,$ from now on let's just see what we can express in terms of 
$a,$ $h,$ $k,$ and $r,$ with the knowledge that whatever we can express in those terms can be expressed in terms of $f$ as well.
Likewise, if $g(x,y)=0$ is the equation of a circle with center $(H,K)$ and radius $R,$ then the most general form of $g$ is
$$g(x,y) = A((x-H)^2 + (y-K)^2 - R^2),$$
where $A,$ $H,$ $K,$ and $R$ can all be expressed in terms of $g.$
Let $\ell$ be the line satisfying the equation
$$ L(x,y) = nx + \left(\sqrt{1-n^2}\right)y - m = 0 $$
where $m$ and $n$ are real constants, $-1 \leq n \leq 1.$
The vector $\left\langle n, \sqrt{1-n^2}\right\rangle$ is a unit normal vector of $\ell.$ (Another way to express this is that $n=\cos\theta$ and 
$\sqrt{1-n^2} = \sin\theta,$ where $\theta$ is the angle of the vector.)
Given any line in the plane, we can select values of $m$ and $n$ such that $\ell$ is the given line.
The distance from a point $(x,y)$ to the line $\ell$ is
$$
d((x,y),\ell) = \left\lvert nx + \left(\sqrt{1-n^2}\right)y - m\right\rvert.
$$
In order to find common tangents to the two circles whose equations are 
$f(x,y)=0$ and $g(x,y)=0,$ therefore, we need to find $m$ and $n$ such that
the distance from $(h,k)$ to $\ell$ is $r$ and the distance from $(H,K)$ to $\ell$ is $R,$ that is,
\begin{align}
nh + \left(\sqrt{1-n^2}\right)k - m &= \pm r, \\
nH + \left(\sqrt{1-n^2}\right)K - m &= \pm R.
\end{align}
If we select the signs of $\pm r$ and $\pm R$ in these two equations,
and solve for $m$ and $n,$ we derive the equation of a line tangent to
the two circles.
There are four possible choices of signs, leading to four tangent
lines whose equations we can derive:
for $+r,+R$ or $-r,-R$ we get a direct tangent,
but for $+r,-R$ or $-r,+R$ we get a transverse tangent.
