Verified Solution • 2025–26 Session moderate

QJEEM24Ma6233: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6233 Solution: Let and be two circles. If the set of all values of so that the …

Let C:x2+y2=4C: x^2+y^2=4 and C′:x2+y2−4λx+9=0C^{\prime}: x^2+y^2-4 \lambda x+9=0 be two circles. If the set of all values of λ\lambda so that the circles C\mathrm{C} and C\mathrm{C} intersect at two distinct points, is R−[a,b]\mathrm{R}-[\mathrm{a}, \mathrm{b}] , then the point (8a+12,16 b−20)(8 \mathrm{a}+12,16 \mathrm{~b}-20) lies on the curve :

Multiple Choice Options

  • A x2+2y2−5x+6y=3x^2+2 y^2-5 x+6 y=3
  • B 5x2−y=−115 x^2-y=-11
  • C x2−4y2=7x^2-4 y^2=7
  • D 6x2+y2=426 x^2+y^2=42

Step-by-Step Verified Solution

amp;C:x2+y2=4⇒C(0,0),r1=2amp;C′:x2+y2−4λx+9=0⇒C′(2λ,0),r2=4λ2−9amp;∣r1−r2∣lt;CC′lt;∣r1+r2∣amp;∣2−4λ2−9∣lt;∣2λ∣lt;2+4λ2−9amp;∣2λ∣gt;∣2−4λ2−9∣amp;⇒4λ2gt;4+4λ2−9−44λ2−9amp;44λ2−9+5gt;0⇒λ∈R\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}

amp;∣2λ∣lt;2+4λ2−9amp;⇒4λ2lt;4+4λ2−9+4(4λ2)−9amp;5lt;44λ2−9 and λ2≥94\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}

amp;2516lt;4λ2−9amp;⇒λ2gt;16964amp;λ∈(−∞,−138)∪(138,∞)amp;λ∈R−[−138,138]amp;a=−138,b=138amp;⇒(8a+12,16b−20)=(−1,6)amp;⇒6(−1)2+(6)2=42\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}

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