A circle C of radius 2 lies in the second qua | vedclass

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Question

(A) The circle $C$ lies in the second quadrant and touches both axes,so its center is $(-2, 2)$ and its radius $R = 2$.The equation of circle $C$ is $

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) The circle $C$ lies in the second quadrant and touches both axes,so its center is $(-2, 2)$ and its radius $R = 2$.The equation of circle $C$ is $(x + 2)^2 + (y - 2)^2 = 2^2 = 4$.The second circle has center $(2, 5)$ and radius $r$. Its equation is $(x - 2)^2 + (y - 5)^2 = r^2$.The distance $d$ between the centers $(-2, 2)$ and $(2, 5)$ is $d = \sqrt{(2 - (-2))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.For two circles to intersect at exactly two points,the distance $d$ between their centers must satisfy $|R - r| Substituting the values,we get $|2 - r| From $5  3$.From $|2 - r| Combining these,we get $3 Thus,the interval $(\alpha, \beta)$ is $(3, 7)$,so $\alpha = 3$ and $\beta = 7$.The value of $3 \beta - 2 \alpha = 3(7) - 2(3) = 21 - 6 = 15$.

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