Aspire Faculty ID #17946 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

Let the set of all values of $r$, for which the circles $(x + 1)^2 + (y + 4)^2 = r^2$ and $x^2 + y^2 - 4x - 2y - 4 = 0$ intersect at two distinct points be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

Solution

$(x-2)^2 + (y-1)^2 = 9$ & $(x+1)^2 + (y+4)^2 = r^2$

$r_1 < c < r_1 + r_2$

$|r-3| < \sqrt{(2+1)^2 + (1+4)^2} < r + 3$

$|r-3| < \sqrt{34} < r + 3$

$-\sqrt{34} < r-3 < \sqrt{34}$

$\Rightarrow r \in (3-\sqrt{34}, 3+\sqrt{34})$

$\therefore \alpha\beta = (3-\sqrt{34})(3+\sqrt{34}) = 9 - 34 = -25$

$\Rightarrow 25$

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