Let C: x ^ 2 + y ^ 2 = 4 and C ^ ' : x ^ 2 + y ^ 2 - 4 + 9 = 0 be...

Given: ${x}^{2} + {y}^{2} = 4$

So, centre and radius of $C$ are \left(0, 0\right)\text{and}{r}_{1} = 2$respectively.</p><p>Also,${C}^{'}: {x}^{2} + {y}^{2} - 4\lambdax + 9 = 0$</p><p>So, centre and radius of${C}^{'} are \left(2\lambda, 0\right)$and${r}_{2} = \sqrt{4{\lambda}^{2} - 9}$respectively.</p><p>$⇒ \left|{r}_{1} - {r}_{2}\right|$< C{C}^{'} < \left|{r}_{1} + {r}_{2}\right|$

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

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

$⇒ \lambda \in R . . . \left(i\right)</p><p>Also,$4{\lambda}^{2} < 4 + 4{\lambda}^{2} - 9 + 4\sqrt{4{\lambda}^{2} - 9}$</p><p>$⇒ 5 < 4\sqrt{4{\lambda}^{2} - 9}\left\{{\lambda}^{2} \ge \dfrac{9}{4}$⇒ \lambda \in \left(- \infty, - \dfrac{3}{2}\right]$ \cup \left[\dfrac{3}{2},\infty\right)</p><p>⇒ \dfrac{25}{16} < 4{\lambda}^{2} - 9$</p><p>$⇒ \dfrac{169}{64} < {\lambda}^{2}$</p><p>$⇒ \dfrac{169}{64} < {\lambda}^{2}$</p><p>$⇒ \lambda \in \left(- \infty, - \dfrac{13}{8}\right)$\cup \left(\dfrac{13}{8}, \infty\right)$ . . . \left(ii\right)$</p><p>Using \left(i\right)\text{and}\left(ii\right)$,

$⇒ \lambda \in \left(- \infty, - \dfrac{13}{8}\right)$ \cup \left(\dfrac{13}{8}, \infty\right)$</p><p>$⇒ \lambda \in R - \left[$-$\dfrac{13}{8}, \dfrac{13}{8}\right]

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

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