1. Equation of the tangent line and distance from center

The given line is $y=2x \quad \Rightarrow \quad 2x-y=0.$

The circle has center $(0,\alpha)$ and radius $r$.

Since the line touches the circle, the perpendicular distance from the center to the line equals the radius: $r=\frac{|2\cdot 0-\alpha|}{\sqrt{2^2+(-1)^2}}=\frac{\alpha}{\sqrt{5}}$ because $\alpha>0$.

So, $r=\frac{\alpha}{\sqrt{5}}.$

2. Use the condition $\alpha+r=5+\sqrt{5}$

Substitute $r=\dfrac{\alpha}{\sqrt{5}}$: $\alpha+\frac{\alpha}{\sqrt{5}}=5+\sqrt{5}.$

Factor out $\alpha$: $\alpha\left(1+\frac{1}{\sqrt{5}}\right)=5+\sqrt{5}.$

Now, $1+\frac{1}{\sqrt{5}}=\frac{\sqrt{5}+1}{\sqrt{5}}.$

Hence, $\alpha\cdot \frac{\sqrt{5}+1}{\sqrt{5}}=5+\sqrt{5}=\sqrt{5}(\sqrt{5}+1).$

Therefore, $\alpha=\frac{\sqrt{5}(\sqrt{5}+1)\cdot \sqrt{5}}{\sqrt{5}+1}=5.$

So, $\boxed{\alpha=5}.$

Then $r=\frac{\alpha}{\sqrt{5}}=\frac{5}{\sqrt{5}}=\sqrt{5}.$

So, $\boxed{r=\sqrt{5}}.$

Thus:

• $(P)\to (4)$ since $\alpha=5$
• $(Q)\to (2)$ since $r=\sqrt{5}$

3. Find the point of contact $A_1$

The tangent line has slope $2$, so the radius to the point of contact is perpendicular to it and has slope $-\frac{1}{2}.$

The radius passes through the center $(0,5)$, so its equation is $y-5=-\frac12(x-0),$ that is, $y=5-\frac{x}{2}.$

The point of contact $A_1$ lies on both this line and the tangent line $y=2x$.

So solve: $2x=5-\frac{x}{2}.$ Multiply by $2$: $4x=10-x$ $5x=10$ $x=2.$ Then $y=2x=4.$

Hence, $\boxed{A_1=(2,4)}.$

So,

• $(R)\to (5)$

4. Find the diametrically opposite point $B_1$

The center is the midpoint of diameter $A_1B_1$. Let $B_1=(x,y).$ Since midpoint of $A_1=(2,4)$ and $B_1=(x,y)$ is $(0,5)$, $\frac{2+x}{2}=0, \qquad \frac{4+y}{2}=5.$

From these, $2+x=0 \Rightarrow x=-2,$ $4+y=10 \Rightarrow y=6.$

Thus, $\boxed{B_1=(-2,6)}.$

So,

• $(S)\to (3)$

5. Final matching

Therefore the correct matching is: $(P)\to(4), \quad (Q)\to(2), \quad (R)\to(5), \quad (S)\to(3).$

This corresponds to Option C.