Question 15

The circle has center $C(0, \alpha)$ and radius $r$. The line $y = 2x$, or $2x - y = 0$, is tangent to the circle at $A_1$.

1. Find $\alpha$ and $r$: The perpendicular distance from the center $(0, \alpha)$ to the line $2x - y = 0$ is equal to the radius $r$. $r = \frac{|2(0) - \alpha|}{\sqrt{2^2 + (-1)^2}} = \frac{|-\alpha|}{\sqrt{5}}$ Given $\alpha > 0$, we have $r = \frac{\alpha}{\sqrt{5}}$ or $\alpha = r\sqrt{5}$.

Using the given equation $\alpha + r = 5 + \sqrt{5}$: $r\sqrt{5} + r = 5 + \sqrt{5}$ $r(\sqrt{5} + 1) = \sqrt{5}(\sqrt{5} + 1)$ $\Rightarrow r = \sqrt{5}$ $\Rightarrow \alpha = (\sqrt{5})(\sqrt{5}) = 5$ So, (P) matches (4) and (Q) matches (2).

2. Find $A_1$: $A_1$ is the foot of the perpendicular from $C(0, 5)$ to the line $y = 2x$. The line $CA_1$ is perpendicular to $y = 2x$, so its slope is $-\frac{1}{2}$. Equation of $CA_1$: $y - 5 = -\frac{1}{2}(x - 0) \Rightarrow x + 2y = 10$. Intersection of $y = 2x$ and $x + 2y = 10$: $x + 2(2x) = 10 \Rightarrow 5x = 10 \Rightarrow x = 2$ $y = 2(2) = 4$ So, $A_1 = (2, 4)$. Thus, (R) matches (5).

3. Find $B_1$: $B_1$ is the point such that $A_1 B_1$ is a diameter. The center $C(0, 5)$ is the midpoint of $A_1 B_1$. Let $B_1 = (x_1, y_1)$. $\frac{2 + x_1}{2} = 0 \Rightarrow x_1 = -2$ $\frac{4 + y_1}{2} = 5 \Rightarrow y_1 = 6$ So, $B_1 = (-2, 6)$. Thus, (S) matches (3).

Conclusion: P-4, Q-2, R-5, S-3. Correct Option: C