JEE Main 2024 — Circle Question with Solution

Given: $x^2+y^2=4$

So, centre and radius of $C$ are $\left(0,0\right) \text{and} r_1=2$ respectively.

Also, $C^{\text{'}}:x^2+y^2-4\lambda x+9=0$

So, centre and radius of $C^{\text{'}}$ are $\left(2\lambda ,0\right)$ and $r_2=\sqrt{4{\lambda }^2-9}$ respectively.

$\Rightarrow \left|r_1-r_2\right|<C{C}^{\text{'}}<\left|r_1+r_2\right|$

$\Rightarrow \left|2-\sqrt{4{\lambda }^2-9}\right|<\left|2\lambda \right|<2+\sqrt{4{\lambda }^2-9}$

$\Rightarrow 4+4{\lambda }^2-9-4\sqrt{4{\lambda }^2-9}<4{\lambda }^2$

$\Rightarrow \lambda \in R ...\left(i\right)$

Also, $4{\lambda }^2<4+4{\lambda }^2-9+4\sqrt{4{\lambda }^2-9}$

$\Rightarrow 5<4\sqrt{4{\lambda }^2-9} \left\{{\lambda }^2\ge \frac{9}{4}\Rightarrow \lambda \in \left(-\infty ,-\frac{3}{2}\right]\cup \left[\frac{3}{2},\infty \right)\right\}$

$\Rightarrow \frac{25}{16}<4{\lambda }^2-9$

$\Rightarrow \frac{169}{64}<{\lambda }^2$

$\Rightarrow \frac{169}{64}<{\lambda }^2$

$\Rightarrow \lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right) ...\left(ii\right)$

Using $\left(i\right) \text{and} \left(ii\right)$,

$\Rightarrow \lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right)$

$\Rightarrow \lambda \in R-\left[-\frac{13}{8},\frac{13}{8}\right]$

As per question $a=-\frac{13}{8}$ and $b=\frac{13}{8}$

$\therefore$ required point is $\left(8a+12, 16b-20\right)\equiv \left(-1, 6\right)$ with satisfies option $\left(d\right)$.