Let C: x^2+y^2=4 and C^{prime}: x^2+y^2-4 lam | vedclass

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Question

$x^2+y^2=4$

$C(0,0) \quad \quad r_1=2$

$C^{\prime}(2 \lambda, 0) \quad r_2=\sqrt{4 \lambda^2-9}$

$\left|\mathrm{r}_1-\mathrm{r}_2\right|<\

Vedclass · Accepted Answer

$6 x^2+y^2=42$

Vedclass · Answer

$x^2+y^2=4$

$C(0,0) \quad \quad r_1=2$

$C^{\prime}(2 \lambda, 0) \quad r_2=\sqrt{4 \lambda^2-9}$

$\left|\mathrm{r}_1-\mathrm{r}_2ight|<\mathrm{CC} \mathrm{C}^{\prime}<\left|\mathrm{r}_1+\mathrm{r}_2ight|$

$\left|2-\sqrt{4 \lambda^2-9}ight|<|2 \lambda|<2+\sqrt{4 \lambda^2-9}$

$4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9}<4 \lambda^2$

True$\lambda \in$ R.... $(1)$

$4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{4 \lambda^2-9}$

$5<4 \sqrt{4 \lambda^2-9} 	ext { and } \quad \lambda^2 \geq \frac{9}{4}$

$\frac{25}{16}<4 \lambda^2-9 \quad \lambda \in\left(-\infty,-\frac{3}{2}ight] \cup\left[\frac{3}{2}, \inftyight)$

$\frac{169}{64}<\lambda^2$

$\lambda \in\left(-\infty,-\frac{13}{8}ight) \cup\left(\frac{13}{8}, \inftyight)$       $...(2)$

from $(1) $and $(2)$$\lambda \in$

$\lambda \in\left(-\infty,-\frac{13}{8}ight) \cup\left(\frac{13}{8}, \inftyight) \Rightarrow R-\left[-\frac{13}{8}, \frac{13}{8}ight]$

as per question $a=-\frac{13}{8}$ and $b=\frac{13}{8}$

$	herefore \quad$ required point is $(-1,6)$ with satisfies option $(4)$

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 $\mathrm{C}^{\prime}$ intersect at two distinct points, is ${R}-[a, b]$, then the point $(8 a+12,16 b-20)$ lies on the curve:

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