Coordinate Geometry – Circles – JEE Main 22 Jan 2026 Shift 1

Solution:

First circle $C_1$: Center $O_1(-1, -4)$, Radius $r_1 = r$.
Second circle $C_2$: $x^2 + y^2 – 4x – 2y – 4 = 0$.
Center $O_2(2, 1)$, Radius $r_2 = \sqrt{2^2 + 1^2 – (-4)} = \sqrt{4 + 1 + 4} = 3$.

Distance between centers $d = O_1O_2$:
$$ d = \sqrt{(2 – (-1))^2 + (1 – (-4))^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} $$

For two circles to intersect at two distinct points, the condition is:
$$ |r_1 – r_2| < d < r_1 + r_2 $$ $$ |r - 3| < \sqrt{34} < r + 3 $$
Breaking into two inequalities:
1) $\sqrt{34} < r + 3 \Rightarrow r > \sqrt{34} – 3$
2) $|r – 3| < \sqrt{34} \Rightarrow -\sqrt{34} < r - 3 < \sqrt{34} \Rightarrow 3 - \sqrt{34} < r < 3 + \sqrt{34}$
Combining the inequalities for $r > 0$:
$$ (\sqrt{34} – 3) < r < (3 + \sqrt{34}) $$ So, the interval is $(\alpha, \beta) = (\sqrt{34} - 3, \sqrt{34} + 3)$.
We need to find $\alpha\beta$:
$$ \alpha\beta = (\sqrt{34} – 3)(\sqrt{34} + 3) $$
Using $(a-b)(a+b) = a^2 – b^2$:
$$ \alpha\beta = (\sqrt{34})^2 – (3)^2 $$
$$ \alpha\beta = 34 – 9 = 25 $$

Ans. (1)