Question:

Step 1: Determine the condition for maximum angle. For a fixed base $AB$ and a variable point $C$ on a line (in this case, the x-axis), the angle $\angle ACB$ is maximized when the circle passing through points $A$, $B$, and $C$ is tangent to the line at point $C$. Step 2: Apply the Power of a Point theorem. Let the origin be $O(0,0)$. Since the circle passes through $A(0,a)$ and $B(0,\beta)$ and touches the x-axis at $C(x,0)$, the line segment $AB$ lies on the y-axis. The y-axis acts as a secant from the origin $O$ to the circle, and the x-axis acts as a tangent from the origin $O$ to the circle at point $C$. According to the Power of a Point theorem, the square of the tangent segment length from $O$ to $C$ is equal to the product of the lengths of the secant segments from $O$ to $A$ and $O$ to $B$. This means: $$OC^2 = OA \cdot OB$$ Step 3: Calculate the lengths $OA$, $OB$, and $OC$. The coordinates are $A(0,a)$, $B(0,\beta)$, $C(x,0)$, and $O(0,0)$. Since $a, \beta > 0$: The distance $OA$ is the distance from $(0,0)$ to $(0,a)$, which is $a$. $$OA = a$$ The distance $OB$ is the distance from $(0,0)$ to $(0,\beta)$, which is $\beta$. $$OB = \beta$$ The distance $OC$ is the distance from $(0,0)$ to $(x,0)$, which is $|x|$. $$OC = |x|$$ Step 4: Substitute the lengths into the Power of a Point equation and solve for $x$. Substitute the calculated lengths into the equation $OC^2 = OA \cdot OB$: $$(|x|)^2 = a \cdot \beta$$ $$x^2 = a\beta$$ Taking the square root of both sides, we get: $$x = \sqrt{a\beta}$$ Since the problem asks for the value of $x$, and $a, \beta > 0$, we consider the positive square root. The final answer is $\boxed{\sqrt{a\beta}}$.