A circle C_1 of radius 2 touches both x-axis | vedclass

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(B) Let the radius of circle $C_2$ be $r$. Since $C_2$ touches both axes in the first quadrant,its center is $(r, r)$.The distance from the origin $O(

Vedclass · Accepted Answer

$6 + 4 \sqrt{2}$

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(B) Let the radius of circle $C_2$ be $r$. Since $C_2$ touches both axes in the first quadrant,its center is $(r, r)$.The distance from the origin $O(0, 0)$ to the center $A(r, r)$ is $OA = \sqrt{r^2 + r^2} = r\sqrt{2}$.The circle $C_1$ has radius $2$,so its center is $(2, 2)$. The distance from the origin to the center of $C_1$ is $OC = \sqrt{2^2 + 2^2} = 2\sqrt{2}$.Since the circles touch externally,the distance between their centers is the sum of their radii: $OA = OC + 2 + r$.Thus,$r\sqrt{2} = 2\sqrt{2} + 2 + r$.$r(\sqrt{2} - 1) = 2(\sqrt{2} + 1)$.$r = \frac{2(\sqrt{2} + 1)}{\sqrt{2} - 1} 	imes \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = 2(\sqrt{2} + 1)^2 = 2(2 + 1 + 2\sqrt{2}) = 2(3 + 2\sqrt{2}) = 6 + 4\sqrt{2}$.

$A$ circle $C_1$ of radius $2$ touches both $x$-axis and $y$-axis. Another circle $C_2$ whose radius is greater than $2$ touches circle $C_1$ and both the axes. Then the radius of circle $C_2$ is-

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