JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Circle

Equal intercepts on the axes together with exactly three points of intersection mean the circle passes through the origin with centre on $y=x$, say at $(p,p)$ with $p>0$; its radius is then $\sqrt{p^2+p^2}=p\sqrt{2}$, so $r^2=2p^2$. The distance from $(p,p)$ to the line $x+y=1$ is $\frac{|2p-1|}{\sqrt{2}}$. Using ${\left(\frac{\text{chord}}{2}\right)}^2+d^2=r^2$ with chord $\sqrt{14}$, $\frac{14}{4}+\frac{(2p-1{)}^2}{2}=2p^2$, which simplifies to $7+4p^2-4p+1=4p^2$, i.e. $8-4p=0$ and $p=2$. Hence the square of the radius is $r^2=2p^2=8$.