Circle Question 2
Question 2 - 2024 (01 Feb Shift 1)
Let the line $\mathrm{L}: \sqrt{2} \mathrm{x}+\mathrm{y}=\alpha$ pass through the point of the intersection $\mathrm{P}$ (in the first quadrant) of the circle $x^{2}+y^{2}=3$ and the parabola $x^{2}=2 y$. Let the line $L$ touch two circles $C_{1}$ and $C_{2}$ of equal radius $2 \sqrt{3}$. If the centres $Q_{1}$ and $Q_{2}$ of the circles $C_{1}$ and $C_{2}$ lie on the y-axis, then the square of the area of the triangle $P_{1} Q_{2}$ is equal to
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Answer (72)
Solution
$$ \begin{gathered} x^{2}+y^{2}=3 \text { and } x^{2}=2 y \ y^{2}+2 y+3=0 \Rightarrow(y+3)(y-1)=0 \ y=-3 \text { or } y=1 \ \mathrm{y}=1 \mathrm{x}=\sqrt{2} \Rightarrow P(\sqrt{2}, 1) \end{gathered} $$
$\mathrm{p}$ lies on the line
$\sqrt{2} x+y=\alpha$
$\sqrt{2}(\sqrt{2})+1=\alpha$
$\alpha=3$
For circle $\mathrm{C}_{1}$
$\mathrm{Q}_{1}$ lies on $\mathrm{y}$ axis
Let $\mathrm{Q}_{1}(0, \alpha)$ coordinates
$\mathrm{R}_{1}=2 \sqrt{3}$ (Given
Line $\mathrm{L}$ act as tangent
Apply $\mathrm{P}=\mathrm{r}$ (condition of tangency)
$\Rightarrow\left|\frac{\alpha-3}{\sqrt{3}}\right|=2 \sqrt{3}$
$\Rightarrow|\alpha-3|=6$
$\alpha-3=6$ or $\alpha-3=-6$
$\Rightarrow \alpha=9 \quad \alpha=-3$
$\triangle P Q_{1} Q_{2}=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & 1 & 1 \ 0 & 9 & 1 \ 0 & -3 & 1\end{array}\right|$
$=\frac{1}{2}(\sqrt{2}(12))=6 \sqrt{2}$
$\left(\Delta P Q_{1} Q_{2}\right)^{2}=72$
