Mathematics · Circles

JEE Main 2024 — 1 February, Shift 1 — Question 26

Let the line $L: \sqrt{2} \mathrm{x}+\mathrm{y}=\alpha$ pass through the point of the intersection P (in the first quadrant) of the circle $x^{2}+y^{2}=3$ and the parabola $x^{2}=2 y$ . Let the line $L$ touch two circles $C_{1}$ and $C_{2}$ of equal radius $2 \sqrt{3}$ . If the centres $\mathrm{Q}_{1}$ and $\mathrm{Q}_{2}$ of the circles $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ lie on the $y$ -axis, then the square of the area of the triangle $\mathrm{PQ}_{1} \mathrm{Q}_{2}$ is equal to $\qquad$ .

Answer: 72

Numerical answer — enter this value.

Step-by-step solution

Given circle: $x^{2}+y^{2}=3$ Given parabola: $x^{2}=2y \Rightarrow y=\frac{x^{2}}{2}$

Substitute in the circle:

x^{2}+\left(\frac{x^{2}}{2}\right)^{2}=3

x^{2}+\frac{x^{4}}{4}=3

4x^{2}+x^{4}=12

x^{4}+4x^{2}-12=0

Let $u=x^{2}$ :

u^{2}+4u-12=0

(u-2)(u+6)=0 \Rightarrow u=2

Since the point lies in the first quadrant,

x=\sqrt{2}, \quad y=\frac{2}{2}=1

\therefore P(\sqrt{2},1)

Line:

L:\ \sqrt{2}x+y=\alpha

Passing through $P$ :

\alpha=\sqrt{2}(\sqrt{2})+1=3

\therefore L:\ \sqrt{2}x+y=3

Let the centre of each circle be $Q(0,k)$ .

Radius:

r=2\sqrt{3}

Distance of $Q(0,k)$ from the line:

\frac{|k-3|}{\sqrt{(\sqrt{2})^{2}+1^{2}}} =\frac{|k-3|}{\sqrt{3}}

Since the line touches the circle,

\frac{|k-3|}{\sqrt{3}}=2\sqrt{3}

|k-3|=6

k=9 \quad \text{or} \quad k=-3

Q_{1}(0,9), \quad Q_{2}(0,-3)

Base:

Q_{1}Q_{2}=9-(-3)=12

Height (distance of $P$ from $y$ -axis):

=\sqrt{2}

Area of $\triangle PQ_{1}Q_{2}$ :

=\frac{1}{2}\times 12 \times \sqrt{2}=6\sqrt{2}

Square of the area:

(6\sqrt{2})^{2}=72

\boxed{72}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Mathematics
Chapter: Circles
Topic: System of Two Circles and Common Tangents

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 C and C^prime…

Answer: 72

Step-by-step solution

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