JEE Main 2026 Apr 6 Shift 1, Mathematics Q22: Circle: Chord, Tangent And Normal

JEE Main2026Apr 6, Shift 1Mathematics

Let the centre of the circle $x^{2} + y^{2} + 2 g x + 2 f y + 25 = 0$ be in the first quadrant and lie on the line $2 x - y = 4$ . Let the area of an equilateral triangle inscribed in the circle be $273$ . Then the square of the length of the chord of the circle on the line $x = 1$ is __________.

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Solution

The centre is $C = (- g, - f)$ and lies on $2 x - y = 4$ : $- 2 g + f = 4$ .

An equilateral triangle inscribed in a circle of radius $r$ has area $\frac{3 3}{4} r^{2}$ . Setting this equal to $273$ gives $r^{2} = 36$ , so $r = 6$ .

Also $r^{2} = g^{2} + f^{2} - 25 = 36 \Rightarrow g^{2} + f^{2} = 61$ . Substitute $f = 2 g + 4$ : $g^{2} + (2 g + 4)^{2} = 61 \Rightarrow 5 g^{2} + 16 g - 45 = 0$ , giving $g = - 5$ (taking the centre in the first quadrant). Then $f = - 6$ and $C = (5, 6)$ .

Distance from $C = (5, 6)$ to the line $x = 1$ is $d = 4$ . The half-chord length is $r^{2} - d^{2} = 36 - 16 = 20$ , so chord length $= 220$ .

$∴ (chord)^{2} = 4 \cdot 20 = 80$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q22: Circle: Chord, Tangent And Normal

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