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

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(B) The first circle $C_1$ touches both axes and has a radius $r_1 = 2$. Thus,its center is $(2, 2)$.Let the radius of the second circle $C_2$ be $a$.

Vedclass · Accepted Answer

$6 + 4\sqrt{2}$

Vedclass · Answer

(B) The first circle $C_1$ touches both axes and has a radius $r_1 = 2$. Thus,its center is $(2, 2)$.Let the radius of the second circle $C_2$ be $a$. Since $C_2$ also touches both axes,its center is $(a, a)$.Since $C_2$ touches $C_1$ externally,the distance between their centers must be equal to the sum of their radii:$\sqrt{(a - 2)^2 + (a - 2)^2} = a + 2$$\sqrt{2(a - 2)^2} = a + 2$$\sqrt{2}|a - 2| = a + 2$Since $a > 2$,we have $\sqrt{2}(a - 2) = a + 2$.$a\sqrt{2} - 2\sqrt{2} = a + 2$$a(\sqrt{2} - 1) = 2 + 2\sqrt{2}$$a = \frac{2(\sqrt{2} + 1)}{\sqrt{2} - 1}$Rationalizing the denominator:$a = \frac{2(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = 2(2 + 1 + 2\sqrt{2}) = 2(3 + 2\sqrt{2}) = 6 + 4\sqrt{2}$.

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