JEE Main 2026 — Circle Question with Solution

The equation of the circle is $x^2+y^2+2gx+2fy+25=0$.

The centre of the circle is $(-g,-f)$. Since it lies in the first quadrant, $-g>0$ and $-f>0$, which implies $g<0$ and $f<0$.

The centre lies on the line $2x-y=4$, so:

$-2g-(-f)=4\Rightarrow f=2g+4$

Let the radius of the circle be $R$. The side length $a$ of an equilateral triangle inscribed in the circle is $a=\sqrt{3}R$.

The area of the equilateral triangle is given as $27\sqrt{3}$:

$\frac{\sqrt{3}}{4}{a}^2=27\sqrt{3}\Rightarrow \frac{\sqrt{3}}{4}(3R^2)=27\sqrt{3}\Rightarrow R^2=36$

The radius of the circle is also given by $R^2=g^2+f^2-c$. Here $c=25$, so:

$g^2+f^2-25=36\Rightarrow g^2+f^2=61$

Substituting $f=2g+4$ into the equation:

$g^2+(2g+4{)}^2=61$

$5g^2+16g-45=0$

$(5g-9)(g+5)=0$

Since $g<0$, we get $g=-5$.

Then, $f=2(-5)+4=-6$.

The centre of the circle is $(5,6)$ and its radius is $R=6$.

The distance $d$ from the centre $(5,6)$ to the line $x=1$ is:

$d=|5-1|=4$

The length of the chord $L$ on the line $x=1$ is given by:

$L=2\sqrt{R^2-d^2}=2\sqrt{36-16}=2\sqrt{20}$

The square of the length of the chord is:

$L^2=(2\sqrt{20}{)}^2=4\times 20=80$

Answer: $80$