JEE Advanced 2024 — Paper 1 — MATH-IV solution

JEE Advanced2024Paper 1MATH-IV

Let the straight line $y = 2 x$ touch a circle with center $(0, α)$ , $α > 0$ , and radius $r$ at a point $A_{1}$ . Let $B_{1}$ be the point on the circle such that the line segment $A_{1} B_{1}$ is a diameter of the circle. Let $α + r = 5 + 5$ .

Match each entry in List-I to the correct entry in List-II.

List-I	List-II
(P) $α$ equals	(1) $(- 2, 4)$
(Q) $r$ equals	(2) $5$
(R) $A_{1}$ equals	(3) $(- 2, 6)$
(S) $B_{1}$ equals	(4) 5
	(5) $(2, 4)$

The correct option is

A
(P) $\to$ (4); (Q) $\to$ (2); (R) $\to$ (1); (S) $\to$ (3)
B
(P) $\to$ (2); (Q) $\to$ (4); (R) $\to$ (1); (S) $\to$ (3)
C
(P) $\to$ (4); (Q) $\to$ (2); (R) $\to$ (5); (S) $\to$ (3)
D
(P) $\to$ (2); (Q) $\to$ (4); (R) $\to$ (3); (S) $\to$ (5)

Solution

The line $y = 2 x$ , i.e. $2 x - y = 0$ , is tangent to the circle centered at $(0, α)$ with radius $r$ . The perpendicular distance from $(0, α)$ to the line equals $r$ :

$\frac{∣0 - α ∣}{4 + 1} = r ⟹ α = r 5 (α > 0)$

Substituting into $α + r = 5 + 5$ : $r 5 + r = 5 + 5$ , i.e. $r (5 + 1) = 5 (5 + 1)$ , giving $r = 5$ and $α = 5$ .

Center $C = (0, 5)$ . The point $A_{1}$ is the foot of perpendicular from $C$ on the line $y = 2 x$ . Using $tan θ = 2$ (angle of line with x-axis), $cos θ = \frac{1}{5}$ , $sin θ = \frac{2}{5}$ . The distance from $O$ to $A_{1}$ is $O C^{2} - r^{2} = 25 - 5 = 25$ .

$A_{1} = (25 \cdot \frac{1}{5}, 25 \cdot \frac{2}{5}) = (2, 4)$

Since $C$ is the midpoint of $A_{1} B_{1}$ : $B_{1} = 2 C - A_{1} = (0, 10) - (2, 4) = (- 2, 6)$ .

Thus (P) $\to$ (4), (Q) $\to$ (2), (R) $\to$ (5), (S) $\to$ (3). Option (C).

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