nd{gathered}$$

JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Circle Question 38 English Explanation

$$\begin{aligned} & \theta=\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right) \\ & \sin \theta=\frac{\sqrt{63}}{8} \end{aligned}$$

Area of $$\triangle \mathrm{OAP}=\frac{1}{2} \times \alpha\left(\frac{\beta}{2}\right)=\frac{1}{2} \times 5 \times 4 \sin \theta$$

$$\begin{aligned} \Rightarrow \quad & \alpha \beta=40 \times \frac{\sqrt{63}}{8} \\ & \alpha \beta=5 \times \sqrt{63} \\ & (\alpha \beta)^2=25 \times 63=1575 \end{aligned}$$

19:T435,

$$\begin{gathered} C_1: x^2+y^2=25, C_2:(x-\alpha)^2+y^2=16 \\ 5<\alpha<9 \end{gathered}$$

JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Circle Question 38 Hindi Explanation

$$\begin{aligned} & \theta=\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right) \\ & \sin \theta=\frac{\sqrt{63}}{8} \end{aligned}$$

$$\triangle \mathrm{OAP}$$ का क्षेत्रफल $$=\frac{1}{2} \times \alpha\left(\frac{\beta}{2}\right)=\frac{1}{2} \times 5 \times 4 \sin \theta$$

$$\begin{aligned} \Rightarrow \quad & \alpha \beta=40 \times \frac{\sqrt{63}}{8} \\ & \alpha \beta=5 \times \sqrt{63} \\ & (\alpha \beta)^2=25 \times 63=1575 \end{aligned}$$

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