Question

Let $S$ be a circle whose centre is $C(2,3)$ and touching the $y$-axis. Tangents $OA$ and $OB$ are drawn from the origin $O$ to the circle which touches the circle at $A$ and $B.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{The finite slope of the tangent OA is}& \text{(P)}& \frac{12}{13}\\ \text{(B)}& \text{If}\alpha \text{is the acute angle between the}& \text{(Q)}& \frac{54}{13}\\ & \text{tangents, then the value of}\sin \alpha \text{is}& & \\ \text{(C)}& \text{If}P(x_1,y_1)\text{is any point on the circle,}& \text{(R)}& \frac{5}{12}\\ & \text{then the product of the minimum and the}& & \\ & \text{maximum values of}OP\text{, is}& & \\ \text{(D)}& \text{A line is drawn from}S(4,5)\text{, intersecting}& \text{(S)}& 9\\ & \text{the circle at}M\text{and}N\text{. Then the value of}& & \\ & SM\cdot SN\text{is}& & \\ & & \text{(T)}& 4\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(P)}$
• B. $\text{(A)}\to \text{(P)},\text{(B)}\to \text{(R)}$
• C. $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(S)}$
• D. $\text{(A)}\to \text{(Q)},\text{(B)}\to \text{(P)}$

Answer 1

The correct option is A $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(P)}$

$\left(A\right)$
$S:(x-2{)}^2+(y-3{)}^2=4$
Let $y=mx$ be the tangent.
Then, $\left|\frac{2m-3}{\sqrt{1+m^2}}\right|=2$


$4m^2+9-12m=4m^2+4\Rightarrow m=\frac{5}{12}$
$\text{(A)}\to \text{(R)}$

$\left(B\right)$
$\alpha ={90}^{\circ }-\theta ={90}^{\circ }-{\tan }^{-1}\frac{5}{12}={\cot }^{-1}\frac{5}{12}$
$\therefore \sin \alpha =\sin \left({\cot }^{-1}\frac{5}{12}\right)$
$=\sin \left({\sin }^{-1}\frac{12}{13}\right)$
$=\frac{12}{13}$
$\text{(B)}\to \text{(P)}$

$\left(C\right)$
$O{P}_{max}=OC+r=\sqrt{13}+2O{P}_{min}=OC-r=\sqrt{13}-2$
$\therefore O{P}_{max}\cdot O{P}_{min}=13-4=9$
$\text{(C)}\to \text{(S)}$

$\left(D\right)$

$S{C}^2=(4-2{)}^2+(5-3{)}^2=8$
Now, $S{C}^2-C{T}^2=S{T}^2$
$\Rightarrow S{T}^2=4$
We know $S{T}^2=SM\cdot SN$
$\therefore SM\cdot SN=4$
$\text{(D)}\to \text{(T)}$