JEE Main 2026 — Mathematics Coordinate Geometry

The equation of the circle is $x^2 + y^2 + 2gx + 2fy + 25 = 0$x2+y2+2gx+2fy+25=0.

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

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

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

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

The area of the equilateral triangle is given as $27\sqrt{3}$273​:

$$\dfrac{\sqrt{3}}{4}a^2 = 27\sqrt{3} \Rightarrow \dfrac{\sqrt{3}}{4}(3R^2) = 27\sqrt{3} \Rightarrow R^2 = 36$$43​​a2=273​⇒43​​(3R2)=273​⇒R2=36

The radius of the circle is also given by $R^2 = g^2 + f^2 - c$R2=g2+f2−c. Here $c = 25$c=25, so:

$$g^2 + f^2 - 25 = 36 \Rightarrow g^2 + f^2 = 61$$g2+f2−25=36⇒g2+f2=61

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

$$g^2 + (2g + 4)^2 = 61$$g2+(2g+4)2=61

$$5g^2 + 16g - 45 = 0$$5g2+16g−45=0

$$(5g - 9)(g + 5) = 0$$(5g−9)(g+5)=0

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

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

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

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

$$d = |5 - 1| = 4$$d=∣5−1∣=4

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

$$L = 2\sqrt{R^2 - d^2} = 2\sqrt{36 - 16} = 2\sqrt{20}$$L=2R2−d2​=236−16​=220​

The square of the length of the chord is:

$$L^2 = (2\sqrt{20})^2 = 4 \times 20 = 80$$L2=(220​)2=4×20=80

Answer: $80$80