JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 18)
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 :
$x^2+2 y^2-5 x+6 y=3$
$5 x^2-y=-11$
$x^2-4 y^2=7$
$6 x^2+y^2=42$
Explanation
$\begin{aligned} & C: x^2+y^2=4 \Rightarrow C(0,0), r_1=2 \\\\ & C^{\prime}: x^2+y^2-4 \lambda x+9=0 \Rightarrow C^{\prime}(2 \lambda, 0), r_2=\sqrt{4 \lambda^2-9} \\\\ & \left|r_1-r_2\right| < C C^{\prime} < \left|r_1+r_2\right| \\\\ & \left|2-\sqrt{4 \lambda^2-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & |2 \lambda|>\left|2-\sqrt{4 \lambda^2-9}\right| \\\\ & \Rightarrow 4 \lambda^2 > 4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9} \\\\ & 4 \sqrt{4 \lambda^2-9}+5>0 \Rightarrow \lambda \in R\end{aligned}$
$\begin{aligned} & |2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & \Rightarrow 4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{\left(4 \lambda^2\right)-9} \\\\ & 5<4 \sqrt{4 \lambda^2-9} \text { and } \lambda^2 \geq \frac{9}{4}\end{aligned}$
$\begin{aligned} & \frac{25}{16} < 4 \lambda^2-9 \\\\ & \Rightarrow \lambda^2>\frac{169}{64} \\\\ & \lambda \in\left(-\infty, \frac{-13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \\\\ & \lambda \in R-\left[\frac{-13}{8}, \frac{13}{8}\right] \\\\ & a=\frac{-13}{8}, b=\frac{13}{8} \\\\ & \Rightarrow(8 a+12,16 b-20)=(-1,6) \\\\ & \Rightarrow 6(-1)^2+(6)^2=42\end{aligned}$
$\begin{aligned} & |2 \lambda|<2+\sqrt{4 \lambda^2-9} \\\\ & \Rightarrow 4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{\left(4 \lambda^2\right)-9} \\\\ & 5<4 \sqrt{4 \lambda^2-9} \text { and } \lambda^2 \geq \frac{9}{4}\end{aligned}$
$\begin{aligned} & \frac{25}{16} < 4 \lambda^2-9 \\\\ & \Rightarrow \lambda^2>\frac{169}{64} \\\\ & \lambda \in\left(-\infty, \frac{-13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \\\\ & \lambda \in R-\left[\frac{-13}{8}, \frac{13}{8}\right] \\\\ & a=\frac{-13}{8}, b=\frac{13}{8} \\\\ & \Rightarrow(8 a+12,16 b-20)=(-1,6) \\\\ & \Rightarrow 6(-1)^2+(6)^2=42\end{aligned}$
Comments (0)
