Let the straight line $\mathrm{y}=2 \mathrm{x}$ touch a circle with center $(0, \alpha), \alpha>0$, and…

Let the straight line $\mathrm{y}=2 \mathrm{x}$ touch a circle with center $(0, \alpha), \alpha>0$, and radius $\mathrm{r}$ at a point $\mathrm{A}_1$. Let $\mathrm{B}_1$ be the point on the circle such that the line segment $A_1 B_1$ 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. The correct option is

1. $(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)$
2. $(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)$
3. $(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(3)$
4. $(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)$

Solution

Consider centre as $\mathrm{P}(0, \alpha), \alpha>0$ $\left|\frac{2(0)-\alpha}{\sqrt{5}}\right|=r$ $\begin{aligned}& |-\alpha|=\sqrt{5} r \\& \alpha=\sqrt{5} r \\& \therefore \alpha+r=5+\sqrt{5} \\& \sqrt{5} r+r=\sqrt{5}(\sqrt{5}+1) \\& r=\sqrt{5}, \alpha=5 \\& \therefore P(0,5)\end{aligned}$ Foot of perpendicular from $\mathrm{P}$ to line $2 \mathrm{x}-\mathrm{y}=0$ $\begin{aligned}& \frac{x-0}{2}=\frac{y-5}{-1}=\frac{-(2(0)-5)}{5}=1 \\& x=2, y=4 \quad A_1(2,4)\end{aligned}$ Let $\mathrm{B}(\mathrm{p}, \mathrm{q})$ $\therefore \frac{\mathrm{p}+2}{2}=0, \frac{\mathrm{q}+4}{2}=5$ $\therefore \mathrm{p}=-2, \mathrm{q}=6 \quad \mathrm{~B}(-2,6)$

Asked in: JEE Advanced 2024 (Paper 1)

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