Can you solve it?

Let $O$ be the center of the circle with radius equal to $5$ and with $OB \perp{AC}$ intersecting at point $D.$ Point $D$ divides $OB$ into the lengths: $OD = y, DB = 5-y$ , and it also bisects the chord $AC$ into: $AD = DC = x.$ Focusing on the two right triangles $\triangle{ODC}$ and $\triangle{BDC}$ , we have the following Pythagorean relationships:

$x^2 + y ^2 = 5^2$ (for $\triangle{ODC}$ ),

$x^2 + (5-y)^2 = (2\sqrt{5})^2$ (for $\triangle{BDC}$ )

which yield: $x =4, y = 3.$ The length of chord $AC$ is just equal to $|AC| = 2x = 2(4) = \boxed{8}.$