Circle and Tangent Lines
DAILY_CHALLENGE
Grade None
Question:
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 $A_1B_1$ 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) $(2,4)$
(P)→(4) (Q)→(2) (R)→(1) (S)→(3)
(P)→(2) (Q)→(4) (R)→(1) (S)→(3)
(P)→(4) (Q)→(2) (R)→(5) (S)→(3)
(P)→(2) (Q)→(4) (R)→(3) (S)→(5)
Step-by-Step Solution
Key Concept: Distance from center to tangent line equals radius; foot of perpendicular gives tangent point
Distance from $(0,\alpha)$ to line $2x-y=0$ equals $r$:
$$r=\frac{|2(0)-\alpha|}{\sqrt{5}}=\frac{\alpha}{\sqrt{5}}$$
$\alpha+r=\alpha+\frac{\alpha}{\sqrt{5}}=\alpha\cdot\frac{\sqrt{5}+1}{\sqrt{5}}=5+\sqrt{5}=\sqrt{5}(\sqrt{5}+1)$
$$\Rightarrow\frac{\alpha}{\sqrt{5}}=\sqrt{5}\Rightarrow\alpha=5,\quad r=\sqrt{5}.$$
(P)$\alpha=5$→(4). (Q)$r=\sqrt{5}$→(2).
$A_1$: foot of perpendicular from $(0,5)$ to $y=2x$ (i.e., $2x-y=0$).
Perpendicular through $(0,5)$: slope $-1/2$, so $y=-x/2+5$.
Intersect $y=2x$: $2x=-x/2+5\Rightarrow5x/2=5\Rightarrow x=2,y=4$. $A_1=(2,4)$→(5).
$B_1$: diametrically opposite to $A_1$ through center $(0,5)$:
$B_1=2(0,5)-(2,4)=(-2,6)$→(3).
Correct Answer: C