Let C : x 2 + y 2 = 4 and C ' : x 2 + y 2 − 4 λ x + 9 = 0 be two circles. If the set of all values of λ so…

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 |r_1 - r_2| < C'C' < |r_1 + r_2| $ $ \Rightarrow |2 - \sqrt{4\lambda^2 - 9}| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9} $ $ \Rightarrow 4 + 4\lambda^2 - 9 - 4\sqrt{4\lambda^2 - 9} < 4\lambda^2 $ $ \Rightarrow \lambda \in \mathbb{R} \dots i $ Also, $4\lambda^2 < 4 + 4\lambda^2 - 9 + 4\sqrt{4\lambda^2 - 9}$ $ \Rightarrow 5 < 4\sqrt{4\lambda^2 - 9} \{ \lambda^2 \geq \frac{9}{4} \Rightarrow \lambda \in [-\infty, -\frac{3}{2}] \cup [\frac{3}{2}, \infty] \}$ $ \Rightarrow \frac{25}{16}< 4\lambda^2 - 9 $ $ \Rightarrow \frac{169}{64} < \lambda^2 $ $ \Rightarrow \frac{169}{64}< \lambda^2 $ $ \Rightarrow \lambda \in [-\infty, -\frac{13}{8}] \cup [\frac{13}{8}, \infty] \dots ii $ Using $i$ and $ii$, $ \Rightarrow \lambda \in [-\infty, -\frac{13}{8}] \cup [\frac{13}{8}, \infty] $ $ \Rightarrow \lambda \in \mathbb{R} - [-\frac{13}{8}, \frac{13}{8}$] $ $As per question $a = -\frac{13}{8} and b = \frac{13}{8}$ $ \therefore $required point is $(8a + 12, 16b - 20) \equiv (-1, 6)$ with satisfies option $(d)$.