Let the straight line y = 2x touch a circle with center $(0,\mathrm{\alpha <!--\; \alpha \; -->}),\mathrm{\alpha <!--\; \alpha \; -->}>0$, 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 $\mathrm{\alpha <!--\; \alpha \; -->}+r=5+\sqrt{5}.$

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) √5
(R) A1 equals	(3) (-2,6)
(S) B1 equals	(4) 5
	(5) (2,4)

The correct option is

The correct option is:
(P) → (4) (Q) → (2) (R) �� (5) (S) → (3)

Step-by-step Derivation:

Step 1: Relate the center of the circle, the radius, and the tangent line.
The given circle has its center at $(0,\alpha )$ with $\alpha >0$, and its radius is $r$.
The straight line $y=2x$ (which can be rewritten as $2x-y=0$) touches the circle at point A₁.
Since the line is tangent to the circle, the perpendicular distance from the center of the circle to this line must be equal to the radius $r$.

Using the formula for the perpendicular distance from a point $(x_0,y_0)$ to a line $ax+by+c=0$:
$r=\frac{|2(0)-\alpha |}{\sqrt{2^2+(-1{)}^2}}$
Since $\alpha >0$, we have:
$r=\frac{\alpha }{\sqrt{5}}\Rightarrow \alpha =r\sqrt{5}$

Step 2: Solve for $r$ and $\alpha$.
We are given that:
$\alpha +r=5+\sqrt{5}$
Substituting $\alpha =r\sqrt{5}$ into this equation:
$r\sqrt{5}+r=5+\sqrt{5}$
$r(\sqrt{5}+1)=\sqrt{5}(\sqrt{5}+1)$
Dividing both sides by $(\sqrt{5}+1)$:
$r=\sqrt{5}$
This gives the value of $\alpha$ as:
$\alpha =r\sqrt{5}=\sqrt{5}·\sqrt{5}=5$
Thus, we have:
(P) → (4) (since $\alpha =5$)
(Q) → (2) (since $r=\sqrt{5}$)

Step 3: Find the point of contact A₁.
The center of the circle is $C(0,5)$.
The line joining the center $C$ and the point of contact A₁ is normal to the tangent line $y=2x$.
Since the slope of the tangent line is $2$, the slope of the normal line is $-\frac{1}{2}$.
The equation of the normal passing through $C(0,5)$ is:
$y-5=-\frac{1}{2}(x-0)\Rightarrow x+2y-10=0$

The point of contact A₁ is the intersection of the tangent line $y=2x$ and the normal line $x+2y-10=0$.
Substituting $y=2x$ into the normal equation:
$x+2(2x)-10=0\Rightarrow 5x=10\Rightarrow x=2$
Then, $y=2(2)=4$.
Therefore, the coordinates of A₁ are $(2,4)$.
Thus, (R) → (5).

Step 4: Find the point B₁.
Since A₁B₁ is a diameter of the circle, the center $C(0,5)$ is the midpoint of the segment A₁B₁.
Let the coordinates of B₁ be $(x_1,y_1)$.
Using the midpoint formula:
$\frac{x_1+2}{2}=0\Rightarrow x_1=-2$
$\frac{y_1+4}{2}=5\Rightarrow y_1+4=10\Rightarrow y_1=6$
Therefore, the coordinates of B₁ are $(-2,6)$.
Thus, (S) → (3).