JEE Main 2023 — Circle Question with Solution

A Circle having the centre $\left(2,3\right)$ and radius $4$.

So, the equation of circle will be 

${\left(x-2\right)}^2+{\left(y-3\right)}^2=16$

$\Rightarrow x^2+y^2-4x-6y-3=0$

Now, it is given $x+y=3$ passes through two points $P \& Q$ on circle and tangents are drawn at $P \& Q$ intersect at $S\left(\alpha ,\beta \right)$,

So, $x+y=3$ will be chord of contact.

Now equation of chord of contact will be 

$T_{\left(\alpha ,\beta \right)}=0$

$\Rightarrow \alpha x+\beta y-2\left(x+\alpha \right)-3\left(y+\beta \right)-3=0$

$\Rightarrow x\left(\alpha -2\right)+y\left(\beta -3\right)-2\alpha -3\beta -3=0$

Now comparing above equation with $x+y-3=0$, we get

$\frac{\alpha -2}{1}=\frac{\beta -3}{1}=\frac{2\alpha +3\beta +3}{3}$

Now on solving above relation we get, 

$\alpha =-6 \& \beta =-5$

Hence, the value of 

$4\alpha -7\beta =-24+35=11$