A circle C of radius 2 lies in second quadrant and touches both coordinate axes. Let r be radius of

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10-1.Circle and System of Circles

hard

A circle $C$ of radius $2$ lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has centre at the point $(2,5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of $r$ is the interval $(\alpha, \beta)$, then $3 \beta-2 \alpha$ is equal to :

A$15$

B$14$

C$12$

D$10$

(JEE MAIN-2025)

Solution

image
$S_1:(x+2)^2+(y-2)^2=2^2$
$S_2:(x-2)^2+(y-5)^2=r^2$
Both circle intersect at two points
$\therefore\left|r_1-r_2\right| |r-2|<5<2+r$
$\Rightarrow 3 r \in(3,7)$
$\alpha=3, \beta=7$
$3 \beta-2 \alpha=15$

Standard 11

Mathematics

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Solution

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