Let the line L:√2x+y=α pass through the point of the intersection P(in the first quadrant)of the circle x^2+y^2=3 and the parabola x^2=2y. Let the…

Given: ${x}^{2}+{y}^{2}=3$x2+y2=3 and ${x}^{2}=2y$x2=2y

$$\Rightarrow {y}^{2}+2y-3=0$$⇒y2+2y−3=0

$$\Rightarrow (y+3)(y-1)=0$$⇒(y+3)(y−1)=0

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

$$\Rightarrow y=1$$⇒y=1

$$\Rightarrow x=\sqrt{2}$$⇒x=2​

$$\Rightarrow P\equiv (\sqrt{2},1)$$⇒P≡(2​,1)

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

$$\Rightarrow \sqrt{2}(\sqrt{2})+1=\alpha$$⇒2​(2​)+1=α

$$\Rightarrow \alpha =3$$⇒α=3

For circle ${C}_{1}$C1​, ${Q}_{1}$Q1​ lies on $y-$y−axis.

Let ${Q}_{1}\equiv (0,\beta )$Q1​≡(0,β) and ${R}_{1}=2\sqrt{3}$R1​=23​ (given)

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

$$\Rightarrow P=r$$⇒P=r

$$\Rightarrow |\frac{\beta -3}{\sqrt{3}}|=2\sqrt{3}$$⇒∣3​β−3​∣=23​

$$\Rightarrow |\beta -3|=6$$⇒∣β−3∣=6

$$\Rightarrow \beta -3=6,-6$$⇒β−3=6,−6

$$\Rightarrow \beta =9,-3$$⇒β=9,−3

So, {Q}_{1}\equiv (0,9)&{Q}_{2}\equiv (0,-3)

$$\Rightarrow Ar(\Delta P{Q}_{1}{Q}_{2})=\frac{1}{2}|\begin{matrix}\begin{matrix}\sqrt{2} \\ 0 \\ 0\end{matrix} & \begin{matrix}1 \\ 9 \\ -3\end{matrix} & \begin{matrix}1 \\ 1 \\ 1\end{matrix}\end{matrix}|$$⇒Ar(ΔPQ1​Q2​)=21​∣2​00​​19−3​​111​​∣

$$\Rightarrow Ar(\Delta P{Q}_{1}{Q}_{2})=\frac{1}{2}(\sqrt{2}(12))$$⇒Ar(ΔPQ1​Q2​)=21​(2​(12))

$$\Rightarrow Ar(\Delta P{Q}_{1}{Q}_{2})=6\sqrt{2}$$⇒Ar(ΔPQ1​Q2​)=62​

$$\Rightarrow {(\Delta P{Q}_{1}{Q}_{2})}^{2}=72$$⇒(ΔPQ1​Q2​)2=72