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

Accepted Answer

$\begin{gathered} C:{x}^{2}+{y}^{2}=4\Rightarrow C(0,0),{r}_{1}=2 \ {C}^{'}:{x}^{2}+{y}^{2}-4\lambda x+9=0\Rightarrow {C}^{'}(2\lambda ,0),{r}_{2}=\sqrt{4{\lambda }^{2}-9} \ |{r}_{1}-{r}_{2}||2-\sqrt{4{\lambda }^{2}-9}| \ \Rightarrow 4{\lambda }^{2}>4+4{\lambda }^{2}-9-4\sqrt{4{\lambda }^{2}-9} \ 4\sqrt{4{\lambda }^{2}-9}+5>0\Rightarrow \lambda \in R \end{gathered}$

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

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