Circle Question 8
Question 8 - 2024 (30 Jan Shift 2)
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 $\mathrm{C}{1}$ and $\mathrm{C}{2}$ 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
Show Answer
Answer (1575)
Solution
$\mathrm{C}{1}: \mathrm{x}^{2}+\mathrm{y}^{2}=25, \mathrm{C}{2}:(\mathrm{x}-\alpha)^{2}+\mathrm{y}^{2}=16$
$5<\alpha<9$
$\theta=\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right)$
$\sin \theta=\frac{\sqrt{63}}{8}$
Area of $\triangle \mathrm{OAP}=\frac{1}{2} \times \alpha\left(\frac{\beta}{2}\right)=\frac{1}{2} \times 5 \times 4 \sin \theta$
$\Rightarrow \alpha \beta=40 \times \frac{\sqrt{63}}{8}$
$\alpha \beta=5 \times \sqrt{63}$
$(\alpha \beta)^{2}=25 \times 63=1575$
