Match each entry in List-I to the correct entry in List-II and choose the correct option. | List-I | List-II | |--------|--------| | (P) The circle with centre $(1,2)$ and touching the straight line $3 \mathrm{x}+4 \mathrm{y}=1$, passes through | (1) the point $(1,1)$ | | (Q) The common tangent to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=2$ and the parabola $\mathrm{y}^{2}=8 \mathrm{x}$ with positive slope, passes through | (2) the point $(7,9)$ | | (R) Let M be the end point of the latus rectum of the ellipse $3 \mathrm{x}^{2}+4 \mathrm{y}^{2}=48$ such that M lies in the first quadrant. Then the normal to the ellipse drawn at M passes through | (3) the point $(3,2)$ | | (S) Let H be the hyperbola whose centre is at the origin, one of the foci is at (5, 0), and one directrix is $5 x+16=0$. Then H passes through | (4) the point $(2,5)$ | | | (5) the point $(8,3 \sqrt{3})$ |

(B) (P) → (3), (Q) → (2), (R) → (1), (S) → (5)

(A) (P) → (3), (Q) → (4), (R) → (1), (S) → (2)

(C) (P) → (3), (Q) → (2), (R) → (4), (S) → (5)

(D) (P) → (4), (Q) → (1), (R) → (2), (S) → (3)

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Circle

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

List-I	List-II
(P) The circle with centre $(1, 2)$ and touching the straight line $3 x + 4 y = 1$ , passes through	(1) the point $(1, 1)$
(Q) The common tangent to the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = 8 x$ with positive slope, passes through	(2) the point $(7, 9)$
(R) Let M be the end point of the latus rectum of the ellipse $3 x^{2} + 4 y^{2} = 48$ such that M lies in the first quadrant. Then the normal to the ellipse drawn at M passes through	(3) the point $(3, 2)$
(S) Let H be the hyperbola whose centre is at the origin, one of the foci is at (5, 0), and one directrix is $5 x + 16 = 0$ . Then H passes through	(4) the point $(2, 5)$
	(5) the point $(8, 33)$

Answer: (B)

(P) Radius = distance from

(1, 2)

to line

3 x + 4 y = 1

∣3 + 8 - 1∣/5 = 2

. Circle:

(x - 1)^{2} + (y - 2)^{2} = 4

. Point

(3, 2)

satisfies this. P→3.

(Q) Tangent to parabola

y^{2} = 8 x

y = m x + 2/ m

. For circle

x^{2} + y^{2} = 2

, condition gives

m = 1

. Tangent:

y = x + 2

. Point

(7, 9)

satisfies. Q→2.

(R) Ellipse

x^{2} /16 + y^{2} /12 = 1

. Latus rectum endpoint in Q1:

M (2, 3)

. Normal:

2 x - y = 1

. Point

(1, 1)

satisfies. R→1.

(S) From

a e = 5

and

a / e = 16/5

, we get

a = 4, e = 5/4

. Hyperbola:

x^{2} /16 - y^{2} /9 = 1

. Point

(8, 33)

satisfies. S→5.

Answer: (B).

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Single correct option