JEE Main 2026 — Circle Question with Solution

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

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, $c=0$.

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

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

The centre of the circle is $(-g,-f)$. Since it lies in the first quadrant, $-g>0$ and $-f>0$. Therefore, $g=f=-a$ for some $a>0$.

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

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

$d=\frac{|a+a-1|}{\sqrt{1^2+1^2}}=\frac{|2a-1|}{\sqrt{2}}$

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

Squaring both sides, we get:

$4(r^2-d^2)=14\Rightarrow r^2-d^2=\frac{7}{2}$

Substituting $r^2=2a^2$ and $d^2=\frac{(2a-1{)}^2}{2}$:

$2a^2-\frac{(2a-1{)}^2}{2}=\frac{7}{2}$

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

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

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

Answer: $8$