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

Question

Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles $\mathrm{C}$ and $\mathrm{C}$ intersect at two distinct points, is $\mathrm{R}-[\mathrm{a}, \mathrm{b}]$, then the point $(8 \mathrm{a}+12,16 \mathrm{~b}-20)$ lies on the curve :

Accepted Answer

$\begin{aligned} & C: x^2+y^2=4 \Rightarrow C(0,0), r_1=2 \\ & C^{\prime}: x^2+y^2-4 \lambda x+9=0 \Rightarrow C^{\prime}(2 \lambda, 0), r_2=\sqrt{4 \lambda^2-9} \\ & \left|r_1-r_2ight| \left|2-\sqrt{4 \lambda^2-9}ight| \\ & \Rightarrow 4 \lambda^2 > 4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9} \\ & 4 \sqrt{4 \lambda^2-9}+5>0 \Rightarrow \lambda \in R\end{aligned}$

$\begin{aligned} & |2 \lambda|\frac{169}{64} \\ & \lambda \in\left(-\infty, \frac{-13}{8}ight) \cup\left(\frac{13}{8}, \inftyight) \\ & \lambda \in R-\left[\frac{-13}{8}, \frac{13}{8}ight] \\ & a=\frac{-13}{8}, b=\frac{13}{8} \\ & \Rightarrow(8 a+12,16 b-20)=(-1,6) \\ & \Rightarrow 6(-1)^2+(6)^2=42\end{aligned}$

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