JEE Main 2025 Jan 22 Shift 1, Mathematics Q13: Circle: Chord, Tangent And Normal

JEE Main2025Jan 22, Shift 1Mathematics

Let the parabola $y = x^{2} + p x - 3$ , meet the coordinate axes at the points P, Q and R. If the circle C with centre at $(- 1, - 1)$ passes through the points P, Q and R, then the area of $△$ PQR is :

Solution

The y-intercept of the parabola is $R = (0, - 3)$ . The circle has centre $(- 1, - 1)$ and passes through $R$ :

$r^{2} = (0 - (- 1))^{2} + (- 3 - (- 1))^{2} = 1 + 4 = 5$ .

So the circle is $(x + 1)^{2} + (y + 1)^{2} = 5$ . Its x-intercepts (set $y = 0$ ): $(x + 1)^{2} = 4 \Rightarrow x = 1$ or $x = - 3$ . So $P = (1, 0)$ , $Q = (- 3, 0)$ .

Area $= \frac{1}{2} \cdot ∣ P Q ∣ \cdot ∣ y_{R} ∣ = \frac{1}{2} \cdot 4 \cdot 3 = 6$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q13: Circle: Chord, Tangent And Normal

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