Four distinct points \((2 k, 3 k),(1,0),(0,1)\) and \((0,0)\) lie on a circle for \(k\) equal to [JEE Main 2024, 27 Jan …

Given four points \((2 k, 3 k),(1,0),(0,1),(0,0)\)

since \((1,0)\) and \((0,1)\) subtends an angle of \(90^{\circ}\) at \((0,0)\)
Circle passes through \((1,0),(0,1),(0,0)\)

\((x-1)(x-0)+(y-0)(y-1)=0\)
\( x^2+y^2-x-y=0 \)

\((2 k, 3 k)\) lies on circle.
\( \Rightarrow 4 k^2+9 k^2-2 k-3 k=0 \)
\( k=\frac{5}{13}\)