JEE Main 2023 — Circle Question with Solution

Let $S\equiv \left(x_1,y_1\right)$

Given:

$$\begin{gathered} P\equiv \left(-3,2\right) \\ Q\equiv \left(9,10\right) \\ R\equiv \left(\alpha ,4\right) \end{gathered}$$

Since, $PQ\perp RQ$, so

$m_{PQ}\times m_{QR}=-1$

$\Rightarrow \left(\frac{10-2}{9+3}\right)\times \left(\frac{10-4}{9-\alpha }\right)=-1$

$\Rightarrow \left(\frac{1}{9-\alpha }\right)=-\frac{1}{4}$

$\Rightarrow \alpha =13$

So, $R\equiv \left(13,4\right)$

So, centre of circle is $O\equiv \left(\frac{13-3}{2},\frac{4+2}{2}\right)\equiv \left(5,3\right)$

Also,

$m_{OQ}·m_{QS}=-1$

$\Rightarrow \left(\frac{3-10}{5-9}\right)m_{QS}=-1$

$\Rightarrow m_{QS}=-\frac{4}{7}$

Equation of $QS$ is

$y-10=-\frac{4}{7}\left(x-9\right)$

$\Rightarrow 4x+7y=106 \dots \left(1\right)$

Also,

$m_{OR}·m_{RS}=-1$

$\Rightarrow \left(\frac{4-3}{13-5}\right)m_{RS}=-1$

$\Rightarrow m_{RS}=-8$

Equation of $RS$

$y-4=-8\left(x-13\right)$

$\Rightarrow 8x+y=108 \dots \left(2\right)$

Solving equations $\left(1\right) \& \left(2\right)$, we get

$x_1=\frac{25}{2}; y_1=8$

Since, $S\left(x_1,y_1\right)$ lies on $2x-ky=1$

$25-8k=1$

$\Rightarrow 8k=24$

$\Rightarrow \boxed{k=3}$