If the limiting points of the system of circles $x^2+ y^2+ 2gx +w(x^2+ y^2+ 2fy + k)=0$ where $w$ is a parameter , subtends a right angle at origin then find value $k/f^2$?
I know that limiting point denotes the point circle i.e. a circle of radius $0$ but I am not able to use that information to find $k/f^2$?
(See graphics below)
Let us expand the equation of the "linear pencil" of circles, with a division by $1+w$:
$$x^2+y^2+2\frac{g}{1+w} x +2 \frac{fw}{1+w} y +\frac{kw}{1+w}=0$$
that can be written, by "completing the squares":
$$\left(x+\dfrac{g}{1+w}\right)^2-\left(\dfrac{g}{1+w}\right)^2+\left(y+\dfrac{fw}{1+w}\right)^2-\left(\dfrac{fw}{1+w}\right)^2+\dfrac{kw(1+w)}{(1+w)^2}=0$$
or
$$\left(x+\dfrac{g}{1+w}\right)^2+\left(y+\dfrac{fw}{1+w}\right)^2=\dfrac{g^2+f^2w^2-kw(1+w)}{(1+w)^2} \ \ \ (1)$$
By identification with the classical equation of a circle $(x-a)^2+(y-b)^2=R^2$, we deduce that the coordinates of the center $C$ are $$\left(-\dfrac{g}{1+w},-\dfrac{fw}{1+w}\right) \ \ \ (2)$$
Besides, the square of the radius of the circle has to be zero if one considers point-circles, i.e.,
$$(f^2-k)w^2-kw+g^2=0 \ \ \ (3)$$
We can already remark that the product of the roots of (3), considered as a second degree equation in $w$ is:
$$w_1w_2=\dfrac{g^2}{f^2-k} \ \ \ (4)$$
(Thi will prove useful in a short while).
It is time now to to introduce the right angle condition between $\vec{OC_1}$ and $\vec{OC_2}$. It can be written as a null dot product (we take into account the coordinates of point-circles $C_k$ are given by (2)):
$$\left(-\dfrac{g}{1+w_1}\right)\left(-\dfrac{g}{1+w_2}\right) + \left(-\dfrac{fw_1}{1+w_1}\right)\left(-\dfrac{fw_2}{1+w_2}\right)=0\ \ \ (5)$$
(5) yields $$w_1w_2=-\dfrac{g^2}{f^2} \ \ \ (6)$$
This is the product of the roots of equation (3). By identification with (4), we conclude that:
$$\dfrac{k}{f^2}=2 \ \ \ (7)$$
Remark: we have assumed all the way long the existence of two real roots for (3). A posteriori, relationship (7) permits to prove that equation (3) has a positive discriminant.
Graphics in the case $f=2, g=4, k=8$ with $w_1=-3.24, w_2=1.24$.