Circle Question 2
Question 2 - 2024 (01 Feb Shift 1)
Let the line $L: \sqrt{2} x+y=\alpha$ pass through the point of the intersection $P$ (in the first quadrant) of the circle $x^{2}+y^{2}=3$ and the parabola $x^{2}=2 y$. 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 \\ y=1 x=\sqrt{2} \Rightarrow P(\sqrt{2}, 1) \end{gathered} $
$p$ lies on the line
$\sqrt{2} x+y=\alpha$
$\sqrt{2}(\sqrt{2})+1=\alpha$
$\alpha=3$
For circle $C _1$
$Q _1$ lies on $y$ axis
Let $Q _1(0, \alpha)$ coordinates
$R _1=2 \sqrt{3}$ (Given
Line $L$ act as tangent
Apply $P=r$ (condition of tangency)
$\Rightarrow\left|\frac{\alpha-3}{\sqrt{3}}\right|=2 \sqrt{3}$
$\Rightarrow|\alpha-3|=6$
$\alpha-3=6$ or $\alpha-3=-6$
$\Rightarrow \alpha=9 \quad \alpha=-3$
$\triangle P Q _1 Q _2=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1\end{array}\right|$
$=\frac{1}{2}(\sqrt{2}(12))=6 \sqrt{2}$
$\left(\Delta P Q _1 Q _2\right)^{2}=72$
