JEE Main 2024 — 30 January, Shift 1 — Question 14

We are given two circles and the condition that they intersect at exactly two distinct points. Circle 1: $(x + 1)^2 + (y + 2)^2 = r^2$ From this equation, we can identify the center and radius of the first circle: Center $C_1 = (-1, -2)$ Radius $R_1 = r$

Circle 2: $x^2 + y^2 - 4x - 4y + 4 = 0$ To find the center and radius of the second circle, we complete the square: $(x^2 - 4x) + (y^2 - 4y) = -4$ $(x^2 - 4x + 4) + (y^2 - 4y + 4) = -4 + 4 + 4$ $(x - 2)^2 + (y - 2)^2 = 4$ From this equation, we can identify the center and radius of the second circle: Center $C_2 = (2, 2)$ Radius $R_2 = \sqrt{4} = 2$

For two circles to intersect at exactly two distinct points, the distance between their centers ($d$) must satisfy the condition: $|R_1 - R_2| < d < R_1 + R_2$

First, calculate the distance $d$ between the centers $C_1(-1, -2)$ and $C_2(2, 2)$: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $d = \sqrt{(2 - (-1))^2 + (2 - (-2))^2}$ $d = \sqrt{(2 + 1)^2 + (2 + 2)^2}$ $d = \sqrt{(3)^2 + (4)^2}$ $d = \sqrt{9 + 16}$ $d = \sqrt{25}$ $d = 5$

Now, apply the condition for intersection at two distinct points: $|r - 2| < 5 < r + 2$

This inequality can be split into two separate inequalities: Inequality A: $|r - 2| < 5$ This implies $-5 < r - 2 < 5$. Add 2 to all parts of the inequality: $-5 + 2 < r < 5 + 2$ $-3 < r < 7$

Since a radius must be a positive value, $r > 0$. Therefore, combining with $-3 < r < 7$, we get $0 < r < 7$.

Inequality B:}$5 < r + 2$ Subtract 2 from both sides: $5 - 2 < r$ $3 < r$

To satisfy both Inequality A ($0 < r < 7$) and Inequality B ($3 < r$), we need to find the intersection of these two ranges. The common range for $r$ is $3 < r < 7$.

The final answer is $\boxed{3 < r < 7}$.