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

Answer 1

The correct option is C $\left(A\right)\to \left(R\right),\left(B\right)\to \left(U\right)$

$\left(A\right)$
Equation of circle
$(x+5{)}^2+(y-12{)}^2=81$


Distance between the origin and the centre of the circle
$=\sqrt{5^2+{12}^2}=13$
Minimum value of $\sqrt{x^2+y^2}$
$=13-r=13-9=4$
$\left(A\right)\to \left(R\right)$

$\left(B\right)$
$3x+6y=k$
$\Rightarrow \frac{3x+6y}{k}=1\cdots \left(1\right)$
Now, $2x^2+2xy+3y^2-1=0$
To get the equation of lines $OA$ and $OB$, we homogenize the curve using equation $\left(1\right)$.
$\Rightarrow 2x^2+2xy+3y^2-{\left(\frac{3x+6y}{k}\right)}^2=0$


As $AB$ is the diameter of circle, so $OA$ is perpendicular to $OB$.
Therefore, sum of coefficients of $x^2$ and $y^2$ is equal to zero.
$2-\frac{9}{k^2}+3-\frac{36}{k^2}=0\Rightarrow 5-\frac{45}{k^2}=0$
$\therefore k^2=9$
$\left(B\right)\to \left(U\right)$