Let the straight line $$y = 2x$$ touch a circle with center $$(0, \alpha)$$, $$\alpha \gt 0$$, and radius $$r$$ at a point $$A_1$$. Let $$B_1$$ be the point on the circle such that the line segment $$A_1B_1$$ is a diameter of the circle. Let $$\alpha + r = 5 + \sqrt{5}$$.

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)	$$A_1$$ equals	(3)	$$(-2, 6)$$
(S)	$$B_1$$ equals	(4)	5
		(5)	$$(2, 4)$$

The given circle has centre $$(0,\alpha)$$ and radius $$r$$. The line $$y = 2x$$ is tangent to the circle.

Write the line in the form $$Ax + By + C = 0$$: $$y = 2x \;\; \Longrightarrow \;\; 2x - y = 0$$ so $$A = 2,\; B = -1,\; C = 0$$.

Distance of centre $$(0,\alpha)$$ from the line equals the radius: $$\frac{|2\cdot 0 + (-1)\alpha + 0|}{\sqrt{2^{2}+(-1)^{2}}} = r \;\; \Longrightarrow \;\; \frac{\alpha}{\sqrt{5}} = r \quad -(1)$$

Also given: $$\alpha + r = 5 + \sqrt{5} \quad -(2)$$

Substitute $$r = \alpha/\sqrt{5}$$ from $$(1)$$ into $$(2)$$: $$\alpha + \frac{\alpha}{\sqrt{5}} = 5 + \sqrt{5} \;\; \Longrightarrow \;\; \alpha\!\left(1 + \frac{1}{\sqrt{5}}\right) = 5 + \sqrt{5}$$

Multiply numerator and denominator by $$\sqrt{5}$$: $$\alpha = \frac{5 + \sqrt{5}}{1 + 1/\sqrt{5}} = \frac{5 + \sqrt{5}}{(\sqrt{5}+1)/\sqrt{5}} = \sqrt{5}\,\frac{5 + \sqrt{5}}{\sqrt{5}+1} = \frac{5\sqrt{5}+5}{\sqrt{5}+1} = 5$$

Therefore $$\alpha = 5$$ and from $$(1)$$ $$r = \frac{\alpha}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$$

Next find the point of contact $$A_1(x_1,y_1)$$. For a tangent, the radius $$\overrightarrow{OA_1}$$ is normal to the line. A normal vector to $$2x - y = 0$$ is $$(2,-1)$$ whose length is $$\sqrt{5}$$. Unit normal $$\mathbf{u} = \left(\dfrac{2}{\sqrt{5}}, -\dfrac{1}{\sqrt{5}}\right)$$.

Hence $$A_1 = (0,\alpha) + r\,\mathbf{u} = \left(0 + \sqrt{5}\cdot\frac{2}{\sqrt{5}},\; 5 + \sqrt{5}\cdot\left(-\frac{1}{\sqrt{5}}\right)\right) = (2,4)$$

Since $$A_1B_1$$ is a diameter, the centre is the midpoint: $$\overrightarrow{OA_1} = (2,-1) \implies \overrightarrow{OB_1} = -(2,-1) = (-2,1)$$ so $$B_1 = (0,\alpha) + (-2,1) = (-2,6)$$

Collecting the results:
$$\alpha = 5,\quad r = \sqrt{5},\quad A_1 = (2,4),\quad B_1 = (-2,6)$$

Matching with List-II:
(P) $$\alpha$$ $$\to$$ 5 (entry 4)
(Q) $$r$$ $$\to$$ $$\sqrt{5}$$ (entry 2)
(R) $$A_1$$ $$\to$$ $$(2,4)$$ (entry 5)
(S) $$B_1$$ $$\to$$ $$(\!-2,6)$$ (entry 3)

Thus the correct option is:
Option C: (P) $$\to$$ (4), (Q) $$\to$$ (2), (R) $$\to$$ (5), (S) $$\to$$ (3)