Question

A circle $C$ having center at $\left(1 , 2\right)$ and radius equal to $3$ , cuts the members of the family of circles passing through two fixed points $P\left(2 , 6\right)$ and $Q\left(4 , 5\right)$ , such that the common chords pass through a fixed point $\left(x_{1} , y_{1}\right)$ , then the value of $\left(\frac{y_{1} - x_{1}}{7}\right)$ is

Answer 1

Equation of family of circles:
$\left(x - 2\right)\left(x - 4\right)+\left(y - 6\right)\left(y - 5\right)+\lambda \left(x + 2 y - 14\right)=0$
$x^{2}+y^{2}+\left(\lambda - 6\right)x+\left(2 \lambda - 11\right)y+\left(38 - 14 \lambda \right)=0$
Equation of circle ' $C$ ': $x^{2}+y^{2}-2x-4y-4=0$
Equation of common chords:
$\left(\lambda - 4\right)x+\left(2 \lambda - 7\right)y+\left(42 - 14 \lambda \right)=0$
$\lambda \left(x + 2 y - 14\right)-\left(4 x + 7 y - 42\right)=0$
$\therefore $ Fixed point $\left(x_{1} , y_{1}\right)=\left(- 14 , 14\right)$