JEE Main 2024 — Mathematics Coordinate Geometry

Given: ${x}^{2}+{y}^{2}=4$x2+y2=4

So, centre and radius of $C$C are $(0,0)\text{and}{r}_{1}=2$(0,0)andr1​=2 respectively.

Also, ${C}^{'}:{x}^{2}+{y}^{2}-4\lambda x+9=0$C′:x2+y2−4λx+9=0

So, centre and radius of ${C}^{'}$C′ are $(2\lambda ,0)$(2λ,0) and ${r}_{2}=\sqrt{4{\lambda }^{2}-9}$r2​=4λ2−9​ respectively.

$$\Rightarrow |{r}_{1}-{r}_{2}|<C{C}^{'}<|{r}_{1}+{r}_{2}|$$⇒∣r1​−r2​∣<CC′<∣r1​+r2​∣

$$\Rightarrow |2-\sqrt{4{\lambda }^{2}-9}|<|2\lambda |<2+\sqrt{4{\lambda }^{2}-9}$$⇒∣2−4λ2−9​∣<∣2λ∣<2+4λ2−9​

$$\Rightarrow 4+4{\lambda }^{2}-9-4\sqrt{4{\lambda }^{2}-9}<4{\lambda }^{2}$$⇒4+4λ2−9−44λ2−9​<4λ2

$$\Rightarrow \lambda \in R...(i)$$⇒λ∈R...(i)

Also, $4{\lambda }^{2}<4+4{\lambda }^{2}-9+4\sqrt{4{\lambda }^{2}-9}$4λ2<4+4λ2−9+44λ2−9​

$$\Rightarrow 5<4\sqrt{4{\lambda }^{2}-9}{{\lambda }^{2}\geq \frac{9}{4}\Rightarrow \lambda \in (-\infty ,-\frac{3}{2}]\cup [\frac{3}{2},\infty )}$$⇒5<44λ2−9​λ2≥49​⇒λ∈(−∞,−23​]∪[23​,∞)

$$\Rightarrow \frac{25}{16}<4{\lambda }^{2}-9$$⇒1625​<4λ2−9

$$\Rightarrow \frac{169}{64}<{\lambda }^{2}$$⇒64169​<λ2

$$\Rightarrow \frac{169}{64}<{\lambda }^{2}$$⇒64169​<λ2

$$\Rightarrow \lambda \in (-\infty ,-\frac{13}{8})\cup (\frac{13}{8},\infty )...(ii)$$⇒λ∈(−∞,−813​)∪(813​,∞)...(ii)

Using $(i)\text{and}(ii)$(i)and(ii),

$$\Rightarrow \lambda \in (-\infty ,-\frac{13}{8})\cup (\frac{13}{8},\infty )$$⇒λ∈(−∞,−813​)∪(813​,∞)

$$\Rightarrow \lambda \in R-[-\frac{13}{8},\frac{13}{8}]$$⇒λ∈R−[−813​,813​]

As per question $a=-\frac{13}{8}$a=−813​ and $b=\frac{13}{8}$b=813​

$\therefore$∴ required point is $(8a+12,16b-20)\equiv (-1,6)$(8a+12,16b−20)≡(−1,6) with satisfies option $(d)$(d).