JEE Main 2026 — 2 April, Morning Shift — Question 39

Let the center be $(h,k)$ in the first quadrant. Since intercepts on axes are equal, $h=k$. The circle passes through the origin (to have exactly three intersections), so $r^2 = h^2 + k^2 = 2h^2$. Radius $r = \sqrt{2}h$. The chord is on $x+y=1$. Distance from center to line: $d = \frac{|2h-1|}{\sqrt{2}}$. Chord length $= 2\sqrt{r^2 - d^2} = \sqrt{14}$. Substitute: $2\sqrt{2h^2 - \frac{(2h-1)^2}{2}} = \sqrt{14}$. Simplify: $\sqrt{8h-2} = \sqrt{14} \Rightarrow 8h-2=14 \Rightarrow h=2$. Thus $r^2 = 2h^2 = 2\cdot4 = 8$.