Two circles of radii 4 text{ cm} and 1 text{ | vedclass

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(A) Let the two circles with centers $Q$ and $R$ and radii $r_1 = 4$ and $r_2 = 1$ touch each other externally. The distance between their centers is

Vedclass · Accepted Answer

$\frac{24}{25}$

Vedclass · Answer

(A) Let the two circles with centers $Q$ and $R$ and radii $r_1 = 4$ and $r_2 = 1$ touch each other externally. The distance between their centers is $QR = r_1 + r_2 = 4 + 1 = 5$.Let the direct common tangents meet at point $P$. The line joining the centers $Q$ and $R$ passes through $P$.Let $\alpha$ be the angle such that the total angle between the tangents is $2\alpha$. In the right-angled triangle formed by the center of the larger circle,the point of contact,and $P$,we have $\sin \alpha = \frac{r_1 - r_2}{QR} = \frac{4 - 1}{5} = \frac{3}{5}$.Since $\sin \alpha = \frac{3}{5}$,we have $\cos \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$.The angle contained by the tangents is $	heta = 2\alpha$.Therefore,$\sin 	heta = \sin(2\alpha) = 2 \sin \alpha \cos \alpha = 2 	imes \frac{3}{5} 	imes \frac{4}{5} = \frac{24}{25}$.

Two circles of radii $4 \text{ cm}$ and $1 \text{ cm}$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $\sin \theta =$

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