Question

Points $P(-3,2),Q(9,10)$ and $R(\alpha ,4)$ lie on a circle $C$ with $PR$ as its diameter. The tangents to $C$ at the points $Q$ and $R$ intersect at the point $S$. If $S$ lies on the line $2x-ky-1$, then $k$ is equal to_____

Answer 1

Answer: 3

$m_{PQ}\cdot m_{QR}=-1$
$\Rightarrow \frac{10-2}{9+3}\times \frac{10-4}{9-\alpha }=-1\Rightarrow \alpha =13$
$m_{OP}\cdot m_{QS}=-1\Rightarrow m_{QS}=-\frac{4}{7}$

Equation of QS
$y-10=-\frac{4}{7}(x-9)$
$\Rightarrow 4x+7y=106\dots .(1)$
$m_{OR}\cdot m_{RS}=-1\Rightarrow m_{RS}=-8$
Equation of RS
$y-4=-8(x-13)$
$\Rightarrow 8x+y=\mathrm{108...}$(2)
Solving eq. (1) & (2)
$x_1=\frac{25}{2}{y}_1=8$
$S\left(x_1,y_1\right)\text{ lies on }2x-ky=1$
$25-8k=1$
$\Rightarrow 8k=24$
$\Rightarrow k=3$