Question

Let the straight line y = 2x touch a circle with center (0, α), α > 0, and radius r at a point A₁. Let B₁ be the point on the circle such that the line segment A₁B₁ is a diameter of the circle. Let \(α+r=5+\sqrt5\).
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)	\(\sqrt5\)
(R)	A1 equals	(3)	(-2, 6)
(S)	B1 equals	(4)	5
		(5)	(2, 4)

The correct option is

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

Answer 1

The Correct Option is C

Circle and Tangent Problem Solution

1. Equation of the Circle:

The circle has center (0, α) and radius r. Its equation is:

x² + (y − α)² = r². (1)

2. Tangent Condition:

The line y = 2x is tangent to the circle at point A₁. The condition for tangency is:

Perpendicular distance from center (0, α) to the line y = 2x = r.

The perpendicular distance is:

|2(0) − 1(α) + 0| / √(2² + 1²) = r.

Simplifying:

α / √5 = r. (2)

3. Relationship between α and r:

Given α + r = 5 + √5, we substitute:

α + (α / √5) = 5 + √5.

Rearrange:

α(1 + 1/√5) = 5 + √5.

Multiply by √5 on both sides:

α = √5(5 + √5) / (√5 + 1).

Rationalizing the denominator:

α = (5 + √5)(√5 − 1) / (5 − 1).

Simplify:

α = 5, r = √5.

4. Coordinates of A₁:

The point A₁ lies on both the circle and the tangent y = 2x.

Substituting y = 2x into the circle equation:

x² + (2x − 5)² = 5.

Expanding:

x² + 4x² − 20x + 25 = 5.

5x² − 20x + 20 = 0.

Dividing by 5:

(x − 2)² = 0 → x = 2.

Substituting x = 2 into y = 2x:

y = 4.

Thus,

A₁ = (2, 4). (3)

5. Coordinates of B₁:

A₁B₁ is a diameter, so B₁ is the reflection of A₁ about the center (0, 5).

Using the midpoint formula:

( (2 + x) / 2 , (4 + y) / 2 ) = (0, 5).

Solving:

2 + x = 0 → x = -2,

4 + y = 10 → y = 6.

Thus,

B₁ = (-2, 6). (4)

Final Matching:

(P) α = 5 → (4),

(Q) r = √5 → (2),

(R) A₁ = (2, 4) → (5),

(S) B₁ = (-2, 6) → (3).