t{4 \lambda^2-9} \\\\ & \left|r_1-r_2\right| < C C^{\prime} < \left|r_1+r_2\right| \\\\ & \left|2-\sqrt{4 \lambda^2-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & |2 \lambda|>\left|2-\sqrt{4 \lambda^2-9}\right| \\\\ & \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}$

$\begin{aligned} & |2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & \Rightarrow 4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{\left(4 \lambda^2\right)-9} \\\\ & 5<4 \sqrt{4 \lambda^2-9} \text { and } \lambda^2 \geq \frac{9}{4}\end{aligned}$

$\begin{aligned} & \frac{25}{16} < 4 \lambda^2-9 \\\\ & \Rightarrow \lambda^2>\frac{169}{64} \\\\ & \lambda \in\left(-\infty, \frac{-13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \\\\ & \lambda \in R-\left[\frac{-13}{8}, \frac{13}{8}\right] \\\\ & 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}$19:T46b,$\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\right| < C C^{\prime} < \left|r_1+r_2\right| \\\\ & \left|2-\sqrt{4 \lambda^2-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & |2 \lambda|>\left|2-\sqrt{4 \lambda^2-9}\right| \\\\ & \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}$

$\begin{aligned} & |2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & \Rightarrow 4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{\left(4 \lambda^2\right)-9} \\\\ & 5<4 \sqrt{4 \lambda^2-9} \text { और } \lambda^2 \geq \frac{9}{4}\end{aligned}$

$\begin{aligned} & \frac{25}{16} < 4 \lambda^2-9 \\\\ & \Rightarrow \lambda^2>\frac{169}{64} \\\\ & \lambda \in\left(-\infty, \frac{-13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \\\\ & \lambda \in R-\left[\frac{-13}{8}, \frac{13}{8}\right] \\\\ & a=\frac{-13}{8}, b=\frac{13}{8} \\\\ & \Rightarro

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