Question

Let a line $L: 2y + x = 5$ and a circle $C: x^2 + y^2 = 25$ intersect at points $A$ and $B$. Match the entries of List-I with the correct entries of List-II.

List-I	List-II
(P) $OB$ is equal to (where $O$ is the origin)	(1) $\sqrt{80}$
(Q) Length of $AB$, is	(2) 7
(R) $OA$ is equal to (where $O$ is the origin)	(3) $\frac{9}{25}$
(S) $\cos^2 \angle AOB$ is equal to	(4) 5
	(5) $\frac{3}{25}$

• A. (P)→(4); (Q)→(2); (R)→(1); (S)→(5)
• B. (P)→(5); (Q)→(3); (R)→(1); (S)→(3)
• C. (P)→(2); (Q)→(3); (R)→(1); (S)→(5)
• D. (P)→(4); (Q)→(1); (R)→(4); (S)→(3)

Answer 1

Answer: (D)

(P)→(4); (Q)→(1); (R)→(4); (S)→(3)

Answer 2

The circle $C: x^2 + y^2 = 25$ has its center at the origin $O(0,0)$ and radius $R=5$. The line is $L: x + 2y - 5 = 0$.
(P) and (R): Since $A$ and $B$ lie on the circle centered at $O$, $OA = OB = R = 5$. This matches entry (4).
(Q): The distance $d$ from $O(0,0)$ to $x+2y-5=0$ is $d = \frac{|0+2(0)-5|}{\sqrt{1^2+2^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$. The length of the chord $AB = 2\sqrt{R^2 - d^2} = 2\sqrt{5^2 - (\sqrt{5})^2} = 2\sqrt{25-5} = 2\sqrt{20} = 4\sqrt{5} = \sqrt{16 \times 5} = \sqrt{80}$. This matches entry (1).
(S): Using the Law of Cosines in $\triangle AOB$: $AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(\angle AOB)$. $80 = 5^2 + 5^2 - 2(5)(5)\cos(\angle AOB)$ $80 = 50 - 50\cos(\angle AOB)$ $30 = -50\cos(\angle AOB) \implies \cos(\angle AOB) = -\frac{3}{5}$. $\cos^2(\angle AOB) = (-\frac{3}{5})^2 = \frac{9}{25}$. This matches entry (3).