Let the straight line $\mathrm{y}=2 \mathrm{x}$ touch a circle with center $(0, \alpha), \alpha>0$, and radius r at a point $\mathrm{A}_{1}$. Let $\mathrm{B}_{1}$ be the point on the circle such that the line segment $\mathrm{A}_{1} \mathrm{~B}_{1}$ is a diameter of the circle. Let $\alpha+\mathrm{r}=5+\sqrt{5}$. Match each entry in List-I to the correct entry in List-II. | List-I | | List-II | | | --- | --- | --- | --- | | (P) | $\alpha$ equals | (1) | (-2, 4) | | (Q) | r equals | (2) | $\sqrt{5}$ | | (R) | $\mathrm{A}_{1}$ equals | (3) | (-2, 6) | | (S) | $\mathrm{B}_{1}$ equals | (4) | 5 | | | | (5) | $(2,4)$ | The correct option is

(C) $(\mathrm{P}) \rightarrow(4)(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5)(\mathrm{S}) \rightarrow(3)$

(A) $(\mathrm{P}) \rightarrow(4)(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(1)(\mathrm{S}) \rightarrow(3)$

(B) $(\mathrm{P}) \rightarrow(2)(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1)(\mathrm{S}) \rightarrow(3)$

(D) $(\mathrm{P}) \rightarrow(2)(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3)(\mathrm{S}) \rightarrow(5)$

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Circle

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.

The correct option is

Answer: (C)

The line

y = 2 x

(or

2 x - y = 0

) is tangent to the circle with center

P (0, α)

and radius

r

.

Distance from center to line:

r = \frac{∣2 ( 0 ) - α ∣}{5} = \frac{α}{5}

Given

α + r = 5 + 5

α + \frac{α}{5} = 5 + 5

α (1 + \frac{1}{5}) = 5 + 5

α \cdot \frac{5 + 1}{5} = 5 (5 + 1)

α = 5, r = 5

So (P)→(4), (Q)→(2).

The foot of perpendicular from

P (0, 5)

to the line

2 x - y = 0

A_{1}

\frac{x - 0}{2} = \frac{y - 5}{- 1} = \frac{- ( 2 ( 0 ) - 5 )}{5} = 1

x = 2, y = 4

A_{1} = (2, 4)

, which is (5). Thus (R)→(5).

Since

A_{1} B_{1}

is a diameter,

B_{1}

is diametrically opposite to

A_{1}

through center

P (0, 5)

B_{1} = 2 P - A_{1} = 2 (0, 5) - (2, 4) = (- 2, 6)

So (S)→(3).

Correct matching: (P)→(4), (Q)→(2), (R)→(5), (S)→(3), which is option (C).

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option