Thank you for reporting, we will resolve it shortly
Q.
A line is drawn through a fixed point $ P (\alpha , \beta) $ to cut the circle $ x^2 + y ^2 = r^2 $ at $ A $ and $ B $ . Then $ PA \cdot PB $ is equal to
The equation of any line through the point $P(\alpha, \beta)$ is
$\frac{x-\alpha}{\cos \theta}=\frac{y-\beta}{\sin \theta}=k$ (say)
Any point on this line is
$(\alpha+k \,cos\,\theta, \beta+k \,sin\, \theta)$
This point lies on the given circle, if
$(\alpha+k cos \theta)^{2}+(\beta+k \sin \theta)^{2}=r^{2}$
or $k^{2}+2 k(\alpha \cos \theta+\beta \sin \theta)$
$+\alpha^{2}+\beta^{2}-r^{2}=0\, \dots(i)$
Which being quadratic in $k$, gives two values of $k$.
Let $PA =k_{1}, PB=K_{2}$, where $k_{1}, K_{2}$ 2 are the roots of Eq. (i), then
$P A \cdot P B=k_{1} k_{2}=\alpha^{2}+\beta^{2}-r^{2}$