Consider two circles C_1: x^2+y^2=25 and C_2: | vedclass

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(C) Let $O(0, 0)$ be the center of $C_1$ and $A(\alpha, 0)$ be the center of $C_2$. Let $P$ be an intersection point of the two circles. The radii are

Vedclass · Accepted Answer

$1575$

Vedclass · Answer

(C) Let $O(0, 0)$ be the center of $C_1$ and $A(\alpha, 0)$ be the center of $C_2$. Let $P$ be an intersection point of the two circles. The radii are $OP = 5$ and $AP = 4$. The distance between centers is $OA = \alpha$.In $\Delta OAP$,the sides are $5, 4,$ and $\alpha$. The angle at $P$ is $	heta = \sin^{-1}\left(\frac{\sqrt{63}}{8}ight)$,so $\sin 	heta = \frac{\sqrt{63}}{8}$.The area of $\Delta OAP$ can be calculated in two ways:$1$) Area $= \frac{1}{2} 	imes OP 	imes AP 	imes \sin 	heta = \frac{1}{2} 	imes 5 	imes 4 	imes \frac{\sqrt{63}}{8} = \frac{5\sqrt{63}}{4}$.$2$) Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \alpha 	imes \left(\frac{\beta}{2}ight) = \frac{\alpha \beta}{4}$.Equating the two areas: $\frac{\alpha \beta}{4} = \frac{5\sqrt{63}}{4} \Rightarrow \alpha \beta = 5\sqrt{63}$.Therefore,$(\alpha \beta)^2 = 25 	imes 63 = 1575$.

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