JEE Main 2023 — Circle Question with Solution

Given,

The tangents at the points $P$ and $Q$ on the circle $x^2+y^2-2x+y=5$ meet at the point $R\left(\frac{9}{4}, 2\right),$

Now plotting the diagram of the above data we have,

Now we know that,

Length of tangent is given by, $PR=QR=\sqrt{S_1}$

So, $L=\sqrt{S_1}=\sqrt{{\left(\frac{9}{4}\right)}^2+{\left(2\right)}^2-2\times \frac{9}{4}+2-5}$

$\Rightarrow L=\frac{5}{4}$

Now distance between $CR$ will be, $CR=\sqrt{{\left(\frac{9}{4}-1\right)}^2+{\left(2+\frac{1}{2}\right)}^2}=\sqrt{\frac{25}{16}+\frac{25}{4}}=\sqrt{\frac{125}{16}}=\frac{5\sqrt{5}}{4}$

Now finding $PQ$ by equating area, we get

$\frac{1}{2}\times \frac{5}{2}\times L=\frac{{\displaystyle 1}}{{\displaystyle 2}}\times \frac{{\displaystyle PQ}}{2}\times CR$

$\Rightarrow \frac{1}{2}\times \frac{5}{2}\times \frac{5}{4}=\frac{{\displaystyle 1}}{{\displaystyle 2}}\times \frac{{\displaystyle PQ}}{2}\times \frac{5\sqrt{5}}{4}$

$\Rightarrow PQ=\sqrt{5}$

Now finding area of triangle $PQR$ by heron's formula,

$s=\frac{2\sqrt{5}+5}{4}$

So, area will be $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

$\Rightarrow ∆=\sqrt{\frac{2\sqrt{5}+5}{4}\left(\frac{2\sqrt{5}+5}{4}-\frac{5}{4}\right)\left(\frac{2\sqrt{5}+5}{4}-\frac{5}{4}\right)\left(\frac{2\sqrt{5}+5}{4}-\sqrt{5}\right)}$

$\Rightarrow ∆=\sqrt{\frac{2\sqrt{5}+5}{4}\left(\frac{2\sqrt{5}}{4}\right)\left(\frac{2\sqrt{5}}{4}\right)\left(\frac{5-2\sqrt{5}}{4}\right)}$

$\Rightarrow ∆=\sqrt{\left(\frac{5}{4}\right)\left(\frac{25-20}{16}\right)}=\frac{5}{8}$