Question

$$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \begin{array}{cc}& \text{(I) A circle of radius}r\text{passes through the origin having center}\\ & \text{on}y=x\text{. If it intersects the circle}{x}^2+y^2-10x-6y+2=0\\ & \text{orthogonally, then the value of}4\sqrt{2}r\text{is}\end{array}& \text{(P)}0\\ \begin{array}{cc}& \text{(II) If a pair of tangent is drawn from point}P(-5\sqrt{2},-5\sqrt{2})\text{to}\\ & \text{the circle}{x}^2+y^2=25\text{then the angle subtended by the pair of}\\ & \text{tangents to the given circle is}\frac{\pi }{p}\text{then}p\text{is}\end{array}& \text{(Q)}1\\ \begin{array}{cc}& \text{(III) Let A (-4,6) and B be a variable point on the line}|x|=6.\\ & \text{If the distance between}A\text{and}B\text{is atmost}4,\text{then the number}\\ & \text{of integral values of B is}\end{array}& \text{(R)}3\\ \begin{array}{cc}& \text{(IV) The value of}\\ & \underset{n\to \infty }{lim}n\sqrt{(\frac{1}{n}-\tan \frac{1}{n})\sqrt{(\frac{1}{n}-\tan \frac{1}{n})\sqrt{(\frac{1}{n}-\tan \frac{1}{n})...\infty }}}\\ & \text{is}\end{array}& \text{(S)}5\\ & \text{(T)}6\\ & \text{(U)}7\end{array}$$
Which of the following is the only CORRECT combination?

• A. $\left(I\right),\left(S\right)$
• B. $\left(II\right),\left(R\right)$
• C. $\left(III\right),\left(P\right)$
• D. $\left(IV\right),\left(Q\right)$

Answer 1

The correct option is B $\left(II\right),\left(R\right)$

$\left(I\right)$
Let the equation of the circle be - $x^2+y^2+2gx+2fy+c=0$
since, this circle passes through (0,0)
$\Rightarrow$ c = 0 and center = $(-g,-f)$
center lies on curve $y=x$
$\Rightarrow g=f$
Since, circle cuts orthogonally
$\Rightarrow 2g{g}_1+2f{f}_1=c+c_{1}2g(-5)+2f(-3)=0+2\Rightarrow -10g-6g=2\Rightarrow g=\frac{-1}{8}=f\text{Radius of circle,}r=\sqrt{g^2+f^2}=\frac{1}{4\sqrt{2}}$
$\Rightarrow 4\sqrt{2}r=1$

$\left(II\right)$
$\Delta POR$
$OP=\sqrt{(-5\sqrt{2}-0{)}^2+(-5\sqrt{2}-0{)}^2}$
$OP=10$
$\sin \frac{\theta }{2}=\frac{OR}{OP}$
$\sin \frac{\theta }{2}=\frac{5}{10}$
$\frac{\theta }{2}=\frac{\pi }{6}$
$\theta =\frac{\pi }{3}=\frac{\pi }{p}$
$p=3$

$\left(III\right)$
The graph of the function $|x|=6$
Since, $AB\le 4\Rightarrow x=-6$
$\Rightarrow \sqrt{(-6+4{)}^2+(y-6{)}^2}\le 4\Rightarrow (y-6{)}^2\le 12\Rightarrow -\sqrt{12}\le y-6\le \sqrt{12}\Rightarrow \text{Integral values of}y=3,4,5,6,7,8,9$
$\therefore \text{No. of integral values of}B=7$


$\left(IV\right)$
$\text{Replacing}n\text{by}\frac{1}{\theta },\text{we get,}$
$=\underset{\theta \to 0}{lim}\frac{1}{\theta }\left[\sqrt{(\theta -\tan \theta )\sqrt{(\theta -\tan \theta )\sqrt{(\theta -\tan \theta )...\infty }}}\right]$
$=\underset{\theta \to 0}{lim}\frac{1}{\theta }(\theta -\tan \theta {)}^{\frac{1}{2}+\frac{1}{2^2}+...\infty }$
$=\underset{\theta \to 0}{lim}\frac{\theta -\tan \theta }{\theta }$
Applying L'Hospital's Rule,
$=\underset{\theta \to 0}{lim}\frac{1-{\sec }^2\theta }{1}$
$=1-{\sec }^{2}0=0$