JEE Main 2026 — Mathematics Coordinate Geometry

Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$x2+y2+2gx+2fy+c=0.

Since the circle intersects the coordinate axes at exactly three points, it must pass through the origin and intersect each axis at one other distinct point. Thus, $c = 0$c=0.

The lengths of the intercepts on the x-axis and y-axis are $2\sqrt{g^2 - c} = 2|g|$2g2−c​=2∣g∣ and $2\sqrt{f^2 - c} = 2|f|$2f2−c​=2∣f∣.

Given that the intercepts are equal, we have $|g| = |f|$∣g∣=∣f∣.

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. Therefore, $g = f = -a$g=f=−a for some $a > 0$a>0.

The centre is $(a, a)$(a,a) and the radius is $r = \sqrt{a^2 + a^2} = a\sqrt{2}$r=a2+a2​=a2​.

The distance $d$d from the centre $(a, a)$(a,a) to the line $x + y - 1 = 0$x+y−1=0 is given by:

$$d = \dfrac{|a + a - 1|}{\sqrt{1^2 + 1^2}} = \dfrac{|2a - 1|}{\sqrt{2}}$$d=12+12​∣a+a−1∣​=2​∣2a−1∣​

The length of the chord on the given line is $2\sqrt{r^2 - d^2} = \sqrt{14}$2r2−d2​=14​.

Squaring both sides, we get:

$$4(r^2 - d^2) = 14 \Rightarrow r^2 - d^2 = \dfrac{7}{2}$$4(r2−d2)=14⇒r2−d2=27​

Substituting $r^2 = 2a^2$r2=2a2 and $d^2 = \dfrac{(2a - 1)^2}{2}$d2=2(2a−1)2​:

$$2a^2 - \dfrac{(2a - 1)^2}{2} = \dfrac{7}{2}$$2a2−2(2a−1)2​=27​

$$4a^2 - (4a^2 - 4a + 1) = 7$$4a2−(4a2−4a+1)=7

$$4a - 1 = 7 \Rightarrow 4a = 8 \Rightarrow a = 2$$4a−1=7⇒4a=8⇒a=2

The square of the radius of the circle is $r^2 = 2a^2 = 2(2)^2 = 8$r2=2a2=2(2)2=8.

Answer: $8$8