JEE Main 2026 — Jan 22, Shift 1 — Mathematics solution

JEE Main2026Jan 22, Shift 1Mathematics

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

A
25
B
20
C
21
D
24

Solution

Rewrite the second circle: $(x - 2)^{2} + (y - 1)^{2} = 9$ , so centre $C_{2} = (2, 1)$ , radius $r_{2} = 3$ . The first has $C_{1} = (- 1, - 4)$ , radius $r$ .

Distance between centres: $∣ C_{1} C_{2} ∣ = 3^{2} + 5^{2} = 34$ .

For two distinct intersections: $∣ r - 3∣ < 34 < r + 3$ .

From the right inequality: $r > 34 - 3$ .

From the left: $r < 34 + 3$ .

So $r \in (34 - 3, 34 + 3)$ , giving $α β = (34)^{2} - 3^{2} = 34 - 9 = 25$ .

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