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JEE Main 2026 Mathematics question

MathematicsCircleJEE Main 2026

Let the centre of the circle x²+y²+2gx+2fy+25=0 be in the first quadrant and lie on the line 2x−y=4. Let the area of an equilateral triangle inscribed in the circle be 27√3. Then the square of the length of the chord of the circle on the line x=1 is .

Answer: 80

Solution

(√3/4)s²=27√3 gives s²=108, so the circumradius R=s/√3=6. Then g²+f²−25=36, and the centre (−g,−f) satisfies

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Question ID VK-015656