JEE Advanced 2024 Paper 1 · Q15 · Circles
Let the straight line $y = 2x$ touch a circle with center $(0, \alpha)$, $\alpha > 0$, and radius $r$ at a point $A_1$. Let $B_1$ 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.
List-I: (P) $\alpha$ equals (Q) $r$ equals (R) $A_1$ equals (S) $B_1$ equals
List-II: (1) $(-2, 4)$ (2) $\sqrt{5}$ (3) $(-2, 6)$ (4) $5$ (5) $(2, 4)$
Reveal answer + step-by-step solution
Correct answer:C
Solution
Distance from center $(0,\alpha)$ to line $2x-y=0$ is $r$: $\dfrac{|2\cdot 0-\alpha|}{\sqrt{5}}=r\Rightarrow \alpha=r\sqrt{5}$. With $\alpha+r=5+\sqrt{5}$ and $\alpha=r\sqrt{5}$: $r\sqrt{5}+r=5+\sqrt{5}\Rightarrow r(\sqrt{5}+1)=\sqrt{5}(\sqrt{5}+1)\Rightarrow r=\sqrt{5}$, $\alpha=5$. (P)$\to$(4), (Q)$\to$(2). $A_1$ is foot of perpendicular from $(0,5)$ to $y=2x$. Direction of $y=2x$ is $(1,2)/\sqrt{5}$. Project $(0,5)$ onto this line through origin: parameter $t=(0,5)\cdot(1,2)/5=2$. So $A_1=(2,4)$. (R)$\to$(5). $B_1$ is diametrically opposite of $A_1$ through center: $B_1=2(0,5)-(2,4)=(-2,6)$. (S)$\to$(3). Option (C).
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