Mathematics - Circle Question with Solution | TestHub

Given: $x^2+y^2=3$ and $x^2=2y$

$\Rightarrow y^2+2y-3=0$

$\Rightarrow \left(y+3\right)\left(y-1\right)=0$

$\Rightarrow y=1,-3 \left(\text{Rejected as it gives imaginary value of} x\right)$

$\Rightarrow y=1$

$\Rightarrow x=\sqrt{2}$

$\Rightarrow P\equiv \left(\sqrt{2},1\right)$

Now, $P$ lies on the line $\sqrt{2}x+y=\alpha$.

$\Rightarrow \sqrt{2}\left(\sqrt{2}\right)+1=\alpha$

$\Rightarrow \alpha =3$

For circle $C_1$, $Q_1$ lies on $y-$axis.

Let $Q_1\equiv \left(0,\beta \right)$ and $R_1=2\sqrt{3}$ (given)

Line $L$ act as tangent so applying the condition of tangency

$\Rightarrow P=r$

$\Rightarrow \left|\frac{\beta -3}{\sqrt{3}}\right|=2\sqrt{3}$

$\Rightarrow \left|\beta -3\right|=6$

$\Rightarrow \beta -3=6, -6$

$\Rightarrow \beta =9, -3$

So, $Q_1\equiv \left(0,9\right) \& Q_2\equiv \left(0,-3\right)$

$\Rightarrow Ar\left(\Delta P{Q}_1{Q}_2\right)=\frac{1}{2}\left|\begin{array}{ccc}\begin{array}{l}\sqrt{2}\\ 0\\ 0\end{array}& \begin{array}{l}1\\ 9\\ -3\end{array}& \begin{array}{l}1\\ 1\\ 1\end{array}\end{array}\right|$

$\Rightarrow Ar\left(\Delta P{Q}_1{Q}_2\right)=\frac{1}{2}\left(\sqrt{2}\left(12\right)\right)$

$\Rightarrow Ar\left(\Delta P{Q}_1{Q}_2\right)=6\sqrt{2}$

$\Rightarrow {\left(\Delta P{Q}_1{Q}_2\right)}^2=72$