Let the straight line y=2 x touch a circle with centre (0, α), α 0 , and radius r at a point A 1 . Let B 1 be the point on the circle such 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 entries 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:

(P) rightarrow (4) (Q) rightarrow(2) ( R ) rightarrow(5) ( S ) rightarrow(3)

(P) rightarrow(4) ( Q ) rightarrow(2) ( R ) rightarrow(1) ( S ) rightarrow(3)

(P) rightarrow(2) ( Q ) rightarrow(4) ( R ) rightarrow(1) ( S ) rightarrow(3)

(P) rightarrow(2) ( Q ) rightarrow(4) ( R ) rightarrow(3) ( S ) rightarrow(5)

Mathematics · Circles

JEE Advanced 2024 — Paper 1 — Question 15

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

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

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

The correct option is:

Option A:
(P) $\rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)$
Option B:
(P) $\rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)$
Option C:
(P) $\rightarrow$ (4) (Q) $\rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(3)$
Correct
Option D:
(P) $\rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)$

Answer: C

Step-by-step solution

Distance from centre $(0,\alpha)$ to line $y=2x$ equals radius $r$ . $\frac{|2\cdot0-\alpha|}{\sqrt{2^2+(-1)^2}} = r \Rightarrow \frac{\alpha}{\sqrt{5}} = r \Rightarrow \alpha = r\sqrt{5}$ . Given $\alpha + r = 5+\sqrt{5}$ , substitute: $r\sqrt{5}+r = 5+\sqrt{5} \Rightarrow r(\sqrt{5}+1) = \sqrt{5}(\sqrt{5}+1) \Rightarrow r = \sqrt{5}$ . Then $\alpha = r\sqrt{5} = \sqrt{5}\cdot\sqrt{5} = 5$ . Point of tangency $A_1$ is foot of perpendicular from centre to line: $A_1 = \left(\frac{2\alpha}{5},\frac{4\alpha}{5}\right) = \left(\frac{10}{5},\frac{20}{5}\right) = (2,4)$ . Since $A_1B_1$ is a diameter, $B_1$ is the point opposite to $A_1$ with centre as midpoint: $B_1 = (2\cdot0-2,\;2\cdot5-4) = (-2,6)$ . Thus $(P)\to(4),\;(Q)\to(2),\;(R)\to(5),\;(S)\to(3)$ , which matches option C.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Circles
Topic: Tangent & Normal , pair of tangents to circle , chord of contact