JEE Mathematics 2025 Question with Solution

Given: Circle C has radius $2$, lies in the second quadrant, and touches both coordinate axes. The second circle has centre $\left(2, 5\right)$ and radius $r$.

Find: The value of $3\beta - 2\alpha$ where the possible values of $r$ form the interval $\left(\alpha, \beta\right)$.

Since circle C touches both axes in the second quadrant and has radius $2$, its centre is $\left(-2, 2\right)$. Hence the equation of circle C is

The equation of the other circle is

For two circles to intersect at exactly two points, the distance between their centres must satisfy

Here, $r_1 = 2$, $r_2 = r$, and the distance between centres is

So we need

Now solve the inequalities:

which gives

and

which gives

so

Combining with $r > 3$, we get

Therefore,

Now compute

Therefore, the correct option is A.

A circle of radius $2$ tangent to both axes in the second quadrant must have centre at a distance $2$ from each axis, so its centre is $\left(-2, 2\right)$.

Let $A = \left(-2, 2\right)$ and $B = \left(2, 5\right)$. Then

For two distinct intersection points, the distance between centres must be strictly between the difference and the sum of the radii:

From $5 < 2+r$, we get $r > 3$. From $|2-r| < 5$, we get $-3 < r < 7$. Hence

Thus $\alpha = 3$ and $\beta = 7$.

Therefore,

So the answer is $15$.