Circle Question 1
Question 1 - 2024 (01 Feb Shift 1)
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 $\mathbf{R}-[a, b]$, then the point $(8 a+12,16 b-20)$ lies on the curve :
(1) $x^{2}+2 y^{2}-5 x+6 y=3$
(2) $5 x^{2}-y=-11$
(3) $x^{2}-4 y^{2}=7$
(4) $6 x^{2}+y^{2}=42$
Show Answer
Answer (1)
Solution
$x^{2}+y^{2}=4$
$\mathrm{C}(0,0) \quad r_{1}=2$
$C^{\prime}(2 \lambda, 0) \circ r_{2}=\sqrt{4 \lambda^{2}-9}$
$\left|r_{1}-r_{2}\right|<C C^{\prime}<\left|r_{1}+r_{2}\right|$
$\left|2-\sqrt{4 \lambda^{2}-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^{2}-9}$
$4+4 \lambda^{2}-9-4 \sqrt{4 \lambda^{2}-9}<4 \lambda^{2}$
True $\lambda \in \mathrm{R}$…
$4 \lambda^{2}<4+4 \lambda^{2}-9+4 \sqrt{4 \lambda^{2}-9}$
$5<4 \sqrt{4 \lambda^{2}-9}$ and $\quad \lambda^{2} \geq \frac{9}{4}$
$\frac{25}{16}<4 \lambda^{2}-9 \quad \lambda \in\left(-\infty,-\frac{3}{2}\right] \cup\left[\frac{3}{2}, \infty\right)$
$\frac{169}{64}<\lambda^{2}$
$\lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right)$
from (1) and (2)
$\lambda \in$
$\lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \Rightarrow \mathrm{R}-\left[-\frac{13}{8}, \frac{13}{8}\right]$
as per question $a=-\frac{13}{8}$ and $b=\frac{13}{8}$
$\therefore \quad$ required point is $(-1,6)$ with satisfies option (4)
