Question

$$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(A)}& \text{If real numbers}x\text{and}y\text{satisfy}& & \\ & (x+5{)}^2+(y-12{)}^2=81,& & \\ & \text{then the minimum value of}\sqrt{x^2+y^2}\text{is}& \text{(P)}& 1\\ \text{(B)}& \text{The line}3x+6y=k\text{intersects the curve}& & \\ & 2x^2+2xy+3y^2=1\text{at point}A\text{and}B.& & \\ & \text{If the circle with}AB\text{as a diameter passes}& & \\ & \text{through the origin, then the value of}{k}^2\text{is}& \text{(Q)}& 2\\ \text{(C)}& \text{If two perpendicular tangents can be}& & \\ & \text{drawn from the origin to the circle}& & \\ & x^2-6x+y^2-2py+17=0\text{, then}& & \\ & \text{the value of}|p|\text{is}& \text{(R)}& 4\\ \text{(D)}& \text{If the circles}& & \\ & x^2+y^2+(3+\sin \beta )x+\left(2\cos \alpha \right)y=0\text{and}& & \\ & x^2+y^2+\left(2\cos \alpha \right)x+2cy=0\text{touch each}& & \\ & \text{other, then the maximum value of}{}^{\prime }{c}^{\prime }\text{is}& \text{(S)}& 5\\ & & \text{(T)}& 7\\ & & \text{(U)}& 9\end{array}$$
Which of the following is the only CORRECT combination?

• A. $\left(C\right)\to \left(R\right),\left(D\right)\to \left(Q\right)$
• B. $\left(C\right)\to \left(S\right),\left(D\right)\to \left(P\right)$
• C. $\left(C\right)\to \left(P\right),\left(D\right)\to \left(S\right)$
• D. $\left(C\right)\to \left(T\right),\left(D\right)\to \left(P\right)$

Answer 1

The correct option is B $\left(C\right)\to \left(S\right),\left(D\right)\to \left(P\right)$

$\left(C\right)$
Given:
$x^2+y^2-6x-2py+17=0\Rightarrow (x-3{)}^2+(y-p{)}^2=(p^2-8)$
As, the tangents from origin to the given circle are perpendicular to each other,
so origin must lie on the director circle of the given circle.

Equation of director circle,
$(x-3{)}^2+(y-p{)}^2=2(p^2-8)$
Putting $(0,0)$ in the equation of the director circle,
$9+p^2=2p^2-16$
$\Rightarrow p^2=25$
$\Rightarrow p=\pm 5$
From the given options,
$p=5$
$\left(C\right)\to \left(S\right)$

$\left(D\right)$
$x^2+y^2+(3+\sin \beta )x+\left(2\cos \alpha \right)y=0$
$x^2+y^2+\left(2\cos \alpha \right)x+2cy=0$
Both circle touch each other at the origin, so tangent at origin will be same on both circles.
$\frac{(3+\sin \beta )}{2}x+\left(\cos \alpha \right)y=0\cdots \left(1\right)$
$\left(\cos \alpha \right)x+cy=0\cdots \left(2\right)$

Comparing equation $\left(1\right)$ and $\left(2\right)$,
$\frac{3+\sin \beta }{2\cos \alpha }=\frac{\cos \alpha }{c}$
$\Rightarrow c=\frac{2{\cos }^2\alpha }{3+\sin \beta }$
The maximum value of $c$ occurs when $\cos \alpha =1,\sin \beta =-1$.
$\therefore c_{max}=1$
$\left(D\right)\to \left(P\right)$