Tangent to the curve $y = \sqrt x + \sqrt y = c$ (where $c > 0$) at any point on it makes equal intercepts on the axes equal to $c^2$. Hence at the point where this tangent meets the line $y = x$:

Tangent to $\sqrt x + \sqrt y = c$ has the property: sum of x and y intercepts equals $c^2$. The tangent line intersects $y = x$ at one point. By symmetry, that's where the curve crosses $y = x$. Set $y = x$ in original: $2\sqrt x = c$ ⇒ $x = c^2/4$. So point is $(c^2/4, c^2/4)$.