5. In the Cartesian coordinate system, with the point $(1,0)$ as the center and $r$ as the radius, a circle intersects the parabola $y^{2}=x$ at four points $A$, $B, C, D$. If the intersection point $F$ of $A C$ and $B D$ is exactly the focus of the parabola, then $r=$ $\qquad$ .
5. $\frac{\sqrt{15}}{4}$.

Combining the equations of the circle and the parabola, we get
$$
\left\{\begin{array}{l}
y^{2}=x \\
(x-1)^{2}+y^{2}=r^{2}
\end{array}\right.
$$

Eliminating $y$, we get $x^{2}-x+1-r^{2}=0$. From $\Delta=1-4\left(1-r^{2}\right)>0$, we solve to get $r>\frac{\sqrt{3}}{2}$.
According to the problem, quadrilateral $A B C D$ is an isosceles trapezoid, and the intersection point of $A C$ and $B D$ is the focus $F\left(\frac{1}{4}, 0\right)$ of the parabola, so points $A, F, C$ are collinear. Let $A\left(x_{1}, \sqrt{x_{1}}\right), C\left(x_{2},-\sqrt{x_{2}}\right)\left(0\frac{\sqrt{3}}{2}$, which meets the requirements of the problem.