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MathMedAdv 2024 • Paper 1

Match Properties of Circle Tangent to Line and Diameter Endpoints

Let the straight line y=2xy = 2x touch a circle with center (0,α)(0, \alpha), α>0\alpha > 0, and radius rr at a point A1A_1.

Let B1B_1 be the point on the circle such that the line segment A1B1A_1 B_1 is a diameter of the circle. Let α+r=5+5\alpha + r = 5 + \sqrt{5}.

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

List-IList-II(P) α equals(1) (−2,4)(Q) r equals(2) 5(R) A1 equals(3) (−2,6)(S) B1 equals(4) 5(5) (2,4)\begin{array}{ll} \text{\textbf{List-I}} & \text{\textbf{List-II}} \\ \text{(P) } \alpha \text{ equals} & \text{(1) } (-2,4) \\ \text{(Q) } r \text{ equals} & \text{(2) } \sqrt{5} \\ \text{(R) } A_1 \text{ equals} & \text{(3) } (-2,6) \\ \text{(S) } B_1 \text{ equals} & \text{(4) } 5 \\ & \text{(5) } (2,4) \end{array}

The correct option is

Options

A

(P)→(4)(Q)→(2)(R)→(1)(S)→(3)(\text{P}) \rightarrow (4) \quad (\text{Q}) \rightarrow (2) \quad (\text{R}) \rightarrow (1) \quad (\text{S}) \rightarrow (3)

B

(P)→(2)(Q)→(4)(R)→(1)(S)→(3)(\text{P}) \rightarrow (2) \quad (\text{Q}) \rightarrow (4) \quad (\text{R}) \rightarrow (1) \quad (\text{S}) \rightarrow (3)

C

(P)→(4)(Q)→(2)(R)→(5)(S)→(3)(\text{P}) \rightarrow (4) \quad (\text{Q}) \rightarrow (2) \quad (\text{R}) \rightarrow (5) \quad (\text{S}) \rightarrow (3)

Correct
D

(P)→(2)(Q)→(4)(R)→(3)(S)→(5)(\text{P}) \rightarrow (2) \quad (\text{Q}) \rightarrow (4) \quad (\text{R}) \rightarrow (3) \quad (\text{S}) \rightarrow (5)

Topics & Concepts

Conic SectionsCirclesStraight Lines

Step-by-Step Solution

To solve the problem step-by-step:

Step 1: Finding the values of α\alpha and rr

The given line touching the circle is 2x−y=02x - y = 0. The center of the circle is C(0,α)C(0, \alpha) with α>0\alpha > 0, and its radius is rr.

The radius rr is the perpendicular distance from the center C(0,α)C(0, \alpha) to the line 2x−y=02x - y = 0: r=∣2(0)−α∣22+(−1)2=α5r = \frac{|2(0) - \alpha|}{\sqrt{2^2 + (-1)^2}} = \frac{\alpha}{\sqrt{5}}

Since α>0\alpha > 0, we have: α=r5\alpha = r\sqrt{5}

We are given the condition: α+r=5+5\alpha + r = 5 + \sqrt{5}

Substituting α=r5\alpha = r\sqrt{5} into this equation gives: r5+r=5+5r\sqrt{5} + r = 5 + \sqrt{5} r(5+1)=5(5+1)r(\sqrt{5} + 1) = \sqrt{5}(\sqrt{5} + 1)

Since 5+1≠0\sqrt{5} + 1 \neq 0, we divide both sides by (5+1)(\sqrt{5} + 1): r=5r = \sqrt{5}

Now, substituting r=5r = \sqrt{5} back to find α\alpha: α=(5)5=5\alpha = (\sqrt{5})\sqrt{5} = 5

Thus:

  • (P) α=5  ⟹  (P)→(4)\alpha = 5 \implies \mathbf{(P) \rightarrow (4)}
  • (Q) r=5  ⟹  (Q)→(2)r = \sqrt{5} \implies \mathbf{(Q) \rightarrow (2)}

Step 2: Finding the point of contact A1A_1

The point of contact A1A_1 lies on the line y=2xy = 2x. The line joining the center C(0,5)C(0, 5) to A1A_1 is perpendicular to 2x−y=02x - y = 0.

Since the slope of 2x−y=02x - y = 0 is 22, the slope of the line CA1CA_1 is −12-\frac{1}{2}. The equation of the normal line CA1CA_1 passing through C(0,5)C(0, 5) is: y−5=−12(x−0)  ⟹  x+2y=10y - 5 = -\frac{1}{2}(x - 0) \implies x + 2y = 10

To find A1A_1, we solve the system of equations:

  1. y=2xy = 2x
  2. x+2y=10x + 2y = 10

Substituting y=2xy = 2x into the second equation: x+2(2x)=10  ⟹  5x=10  ⟹  x=2x + 2(2x) = 10 \implies 5x = 10 \implies x = 2 y=2(2)=4y = 2(2) = 4

Thus, A1=(2,4)A_1 = (2, 4).

  • (R) A1=(2,4)  ⟹  (R)→(5)A_1 = (2, 4) \implies \mathbf{(R) \rightarrow (5)}

Step 3: Finding the point B1B_1

Since A1B1A_1 B_1 is a diameter of the circle, the center C(0,5)C(0, 5) is the midpoint of the line segment A1B1A_1 B_1.

Let B1=(xB,yB)B_1 = (x_B, y_B). By the midpoint formula: (xB+22,yB+42)=(0,5)\left(\frac{x_B + 2}{2}, \frac{y_B + 4}{2}\right) = (0, 5)

Solving for xBx_B and yBy_B: xB+22=0  ⟹  xB=−2\frac{x_B + 2}{2} = 0 \implies x_B = -2 yB+42=5  ⟹  yB=6\frac{y_B + 4}{2} = 5 \implies y_B = 6

Thus, B1=(−2,6)B_1 = (-2, 6).

  • (S) B1=(−2,6)  ⟹  (S)→(3)B_1 = (-2, 6) \implies \mathbf{(S) \rightarrow (3)}

Conclusion:

The correct matching is: (P)→(4)(Q)→(2)(R)→(5)(S)→(3)(\text{P}) \rightarrow (4) \quad (\text{Q}) \rightarrow (2) \quad (\text{R}) \rightarrow (5) \quad (\text{S}) \rightarrow (3)

This corresponds to Option C.

Match Properties of Circle Tangent to Line and Diameter Endpoints | Mathematics PYQ Solution - JEE Challenger