Here is the problem:-
$L: x – y + 3 = 0$ is the radical line for $S$, the system of coaxial circles. $C: x^2 + y^2 – 2x – 4y – 11 = 0$ is a member of $S$ with $AB$ as the common chord.
(a) Find the equation of $T$, the line of centers of $S$.
(b) Find the equation of $C’ (\in S)$, the circle having $AB$ as diameter.
Part (a) is easy. $T$ is perpendicular to $L$ and passes through $(1, 2)$, the center of $C$.
∴ $ T: x + y – 3 = 0$
To solve part (b), the proper but long way is to solve $C$ and $L$ to find $(x_A, y_A)$ and $(x_B, y_B)$ first. The required equation is then $(x – x_A)(x – x_B) + (y – y_A)(y – y_B) = 0$. The result turns out to be … $C’ : x^2 + y^2 – 6y – 5 = 0$.
But the technique the book used is:-
Equation of $C’$ must be $(x^2 + y^2 – 2x – 4y – 11) + 2k(x – y + 3) = 0 ………….. (*)$
The center of $C’$ is at $([1 – k], [2 + k])$ which is point on $L$.
From which, we find $k =1$.
Putting $k = 1$ back in (*), we obtain the same answer.
The question I want to ask is “where does the $2$ [just before the $k$ in (*)] come from?”
Either an explanation or leaving a source for reference is welcome.
The "$2k$" just stands for "some number". They could have used $k$ instead of $2k$, but then the calculations would have shown that $k=2$. Using $2k$ makes it easier to write down the coordinates of the center of $C'$, and makes these center coordinates simpler.