Question

​​​​​​A circle $C_1$ with centre at the origin meets $x$-axis at $A$ and $B$ (where $A\&B$ lies on negative and positive $x-$axis respectively). Two points $P\left(a\right)$ and $Q\left(b\right)$ are on the circle such that $b-a$ is a constant, where $a$ and $b$ are the parametric angles of the points. $BP$ and $AQ$ meets at $R$. Locus of $R$ is a circle $C_2$.
Let $c$ be the radius of $C_1$ and $d$ be the radius of $C_2$.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \text{For}b-a=\frac{\pi }{2}\text{, the value of}\frac{d^2}{c^2}\text{is}& \text{(P)}& 0\\ \text{(2)}& \text{For}b-a=\frac{\pi }{2}\text{and}c=\sqrt{2}\text{, circle}{C}_2\text{intersects}& \text{(Q)}& 1\\ & \text{the coordinate axes at four points}L,M,N,O.& & \\ & \text{Let the area of the quadrilateral LMNO is}2\sqrt{2}p.& & \\ & \text{Then the value of}p\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ,AP}& \left(R\right)& 2\\ & \text{respectively.}\text{If}{m}_1{m}_2=-1\text{, then}\frac{a}{b}\text{is}& & \\ \text{(4)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ,AP}& \left(S\right)& 3\\ & \text{respectively.}\text{If}{m}_1=m_2\text{, then}\frac{3|b-a|}{\pi }\text{is}& & \\ & & \left(T\right)& 4\end{array}$

Then the CORRECT option is :

• A. $\left(1\right)\to \left(T\right)$
• B. $\left(2\right)\to \left(Q\right)$
• C. $\left(3\right)\to \left(P\right)$
• D. $\left(4\right)\to \left(S\right)$

Answer 1

The correct option is D $\left(4\right)\to \left(S\right)$

Equation of the circle $C_1$ is
$x^2+y^2=c^2$
Coordinates are $P(c\cos a,c\sin a)$ and $Q(c\cos b,c\sin b)$
Equation of $AQ$ is
$y=\frac{c\sin b}{c(\cos b+1)}(x+c)\Rightarrow y=\tan \left(\frac{b}{2}\right)(x+c)\Rightarrow \tan \left(\frac{b}{2}\right)=\frac{y}{x+c}$

Equation of $BP$ is
$y=\frac{c\sin a}{c(\cos a-1)}(x-c)\Rightarrow y=-\cot \left(\frac{a}{2}\right)(x-c)\Rightarrow \tan \left(\frac{a}{2}\right)=-\left(\frac{x-c}{y}\right)$

Let $b-a=2k$
$\therefore \tan k=\frac{\tan \left(\frac{b}{2}\right)-\tan \left(\frac{a}{2}\right)}{1+\tan \left(\frac{b}{2}\right)\tan \left(\frac{a}{2}\right)}\Rightarrow \tan k=\frac{x^2+y^2-c^2}{2cy}\Rightarrow x^2+y^2-2cy\tan k=c^2$ which is the locus of $R$, i.e., the circle $C_2$

For $b-a=\frac{\pi }{2}$,
$\tan k=1$
$\Rightarrow x^2+y^2-2cy=c^2$
$\Rightarrow (x-0{)}^2+(y-c{)}^2=(\sqrt{2}c{)}^2$
Centre of the above circle $(0,c)$ lies on the $y$-axis
and radius, $d=\sqrt{2}c$
$\therefore \frac{d^2}{c^2}=2$
$\left(1\right)\to \left(R\right)$

$x$-intercept of the circle $C_2$ is $(\pm c,0)$
$\therefore \text{area of}LMNO=2\times (\frac{1}{2}\times 2\text{radius}\times c)$
$=2\text{radius}\times c$
$=2\times \sqrt{2}c\times c=2\sqrt{2}{c}^2=4\sqrt{2}$
$\Rightarrow 2\sqrt{2}p=4\sqrt{2}$
$\therefore p=2$
$\left(2\right)\to \left(R\right)$

$m_1=\frac{c\sin b}{c(\cos b-1)}=-\cot \frac{b}{2}{m}_2=\frac{c\sin a}{c(\cos a+1)}=\tan \frac{a}{2}$

If $m_1{m}_2=-1$
then $\cot \frac{b}{2}\tan \frac{a}{2}=1$
$\Rightarrow a=b$
$\left(3\right)\to \left(Q\right)$

If $m_1=m_2$
$\Rightarrow \tan \frac{a}{2}+\cot \frac{b}{2}=0$

$\Rightarrow \frac{\cos |\frac{b}{2}-\frac{a}{2}|}{\cos \left(\frac{b}{2}\right)\sin \left(\frac{a}{2}\right)}=0=\cos \frac{\pi }{2}$

$\Rightarrow |b-a|=\pi$
$\left(4\right)\to \left(S\right)$