Let the straight line $y = 2x$ 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. List-I: (P) α equals (Q) r equals $(R) A_{1}$ equals $(S) B_{1}$ equals List-II: (1) (-2, 4) (2) √5 (3) (-2, 6) (4) 5 (5) (2, 4) The correct option is (1) (P)→(4) (Q)→(2) (R)→(1) (S)→(3) (2) (P)→(2) (Q)→(4) (R)→(1) (S)→(3) (3) (P)→(4) (Q)→(2) (R)→(5) (S)→(3) (4) (P)→(2) (Q)→(4) (R)→(3) (S)→(5)

$\text{Circle: centre } C(0, \alpha),\ \alpha > 0,\ \text{radius } r$ $\text{Line } y = 2x \text{ touches the circle at } A_1$ $A_1B_1 \text{ is a diameter}, \qquad \alpha + r = 5 + \sqrt{5}$ $\text{Find } \alpha,\ r,\ A_1,\ B_1 \text{ and match them}$ $\text{List-I: } (P)\ \alpha \quad (Q)\ r \quad (R)\ A_1 \quad (S)\ B_1$ $x^2 + (y - \alpha)^2 = r^2$ $y = 2x \text{ touches at } A_1 \Rightarrow CA_1 \perp y = 2x$ $d = \dfrac{|2(0) - \alpha|}{\sqrt{2^2 + 1^2}} = r$ $\dfrac{\alpha}{\sqrt{5}} = r$ $\alpha + r = 5 + \sqrt{5}$ $\alpha + \dfrac{\alpha}{\sqrt{5}} = 5 + \sqrt{5}$ $\alpha\left(1 + \dfrac{1}{\sqrt{5}}\right) = 5 + \sqrt{5}$ $\alpha\cdot\dfrac{\sqrt{5} + 1}{\sqrt{5}} = 5 + \sqrt{5}$ $\alpha = \dfrac{\sqrt{5}(5 + \sqrt{5})}{\sqrt{5} + 1} = \dfrac{5(\sqrt{5} + 1)}{\sqrt{5} + 1} = 5$ $r = \dfrac{5}{\sqrt{5}} = \sqrt{5}$ $P \to (4):\ \alpha = 5, \qquad Q \to (2):\ r = \sqrt{5}$ $\text{Foot from } C(0,5) \text{ on } 2x - y = 0$ $A_1 = \left(0 - \dfrac{2(-5)}{5},\ 5 - \dfrac{(-1)(-5)}{5}\right)$ $A_1 = \left(\dfrac{10}{5},\ 5 - 1\right) = (2, 4)$ $R \to (5):\ A_1 = (2, 4)$ $\text{slope } CA_1 = \dfrac{4-5}{2-0} = -\dfrac{1}{2}, \quad \text{slope of line} = 2$ $\left(-\dfrac{1}{2}\right)(2) = -1 \Rightarrow CA_1 \perp \text{line}$ $B_1 = 2C - A_1 = (0, 10) - (2, 4)$ $B_1 = (-2, 6)$ $S \to (3)$ $(P) \to (4),\quad (Q) \to (2),\quad (R) \to (5),\quad (S) \to (3)$ $\boxed{\text{Correct option: } (3)}$