JEE Main 2026 · 6 Apr · Shift 1 · Mathematics Q22

← JEE Main 2026 · 6 Apr · Shift 1

MathematicsInteger+4 / -1Question 22 of 75

Let the centre of the circle $x^2+y^2+2gx+2fy+25=0$ 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 $27\sqrt3$ . Then the square of the length of the chord of the circle on the line $x=1$ is ________.

Your answer

Answer and worked solution

Official answer: 80

Radius. An inscribed equilateral triangle has area $\tfrac{3\sqrt3}4r^2=27\sqrt3$ , so $r^2=36$ .

Centre. $(h,2h-4)$ with $h^2+k^2-25=36$ :
$5h^2-16h-45=0,$
so $h=5$ (first quadrant) and the centre is $(5,6)$ .

Chord on $x=1$ . The line is 4 from the centre, so the chord is $2\sqrt{36-16}$ and its square is 80.

Answer: 80

Topics:Equation of a circle

← Previous question Next question →

JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 22