Question:

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:

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For two circles to intersect at two points, the distance between their centers must be greater than the difference of their radii and less than the sum of their radii.
Updated On: Mar 24, 2025
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The Correct Option is A

Solution and Explanation

The equation of circle C with radius 2 and center \( (-2, 2) \) is: \[ S_1: (x + 2)^2 + (y - 2)^2 = 2^2 \] The equation of the circle with center \( (2, 5) \) and radius \( r \) is: \[ S_2: (x - 2)^2 + (y - 5)^2 = r^2 \] For both circles to intersect at exactly two points, the distance between the centers must satisfy: \[ |r_1 - r_2| <c_1c_2 <r_1 + r_2 \] Where \( r_1 = 2 \) and \( r_2 = r \), the distance between centers is \( 5 \): \[ |r - 2| <5 <r + 2 \] \[ 3 <r <7 \] Thus, \( r \in (3, 7) \). Therefore, \( \alpha = 3 \) and \( \beta = 7 \). \[ 3\beta - 2\alpha = 3 \times 7 - 2 \times 3 = 21 - 6 = 15 \]
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