JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 24)
Consider two circles $$C_1: x^2+y^2=25$$ and $$C_2:(x-\alpha)^2+y^2=16$$, where $$\alpha \in(5,9)$$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $$C_1$$ and $$C_2$$ be $$\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right)$$. If the length of common chord of $$C_1$$ and $$C_2$$ is $$\beta$$, then the value of $$(\alpha \beta)^2$$ equals _______.
Answer
1575
Explanation
$$\begin{gathered} C_1: x^2+y^2=25, C_2:(x-\alpha)^2+y^2=16 \\ 5<\alpha<9 \end{gathered}$$
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$$\begin{aligned} & \theta=\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right) \\ & \sin \theta=\frac{\sqrt{63}}{8} \end{aligned}$$
Area of $$\triangle \mathrm{OAP}=\frac{1}{2} \times \alpha\left(\frac{\beta}{2}\right)=\frac{1}{2} \times 5 \times 4 \sin \theta$$
$$\begin{aligned} \Rightarrow \quad & \alpha \beta=40 \times \frac{\sqrt{63}}{8} \\ & \alpha \beta=5 \times \sqrt{63} \\ & (\alpha \beta)^2=25 \times 63=1575 \end{aligned}$$
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