JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Circle

The centre $(-g,-f)$ lies on $2x-y=4$, so $-2g+f=4$. An equilateral triangle inscribed in a circle of radius $r$ has area $\frac{3\sqrt{3}}{4}{r}^2$; setting this to $27\sqrt{3}$ gives $r^2=36$. With $r^2=g^2+f^2-25=36$, so $g^2+f^2=61$, and $f=2g+4$, substitution gives $5g^2+16g-45=0$, i.e. $(g+5)(5g-9)=0$. The first-quadrant centre requires $g=-5$, $f=-6$, so the centre is $(5,6)$ and $r=6$. The distance from $(5,6)$ to the line $x=1$ is $4$, so the chord on $x=1$ has half-length $\sqrt{36-16}=\sqrt{20}$, and the square of its length is $(2\sqrt{20}{)}^2=80$.