Quadrilateral in quarter circle

You're correct the radius is not unique. Instead, it can be anything within a range of values. To see why, consider my diagram above. First, draw any line segment $AB$ of length $8$. Have $C$ be its midpoint. Consider the line segment $CD$ of length $4$. Note that for any circle circumscribing $\triangle ABD$ that $AB$ would its diameter, so $\measuredangle ADB = 90^{\circ}$. You can also see this because, by isosceles triangles, you have $\measuredangle CAD = \measuredangle CDA$ and $\measuredangle CDB = \measuredangle CBD$, so since the total sum of these angles is $180^{\circ}$, you have that $\measuredangle BDA$ is half of that, i.e., $90^{\circ}$. Regardless of how you determine it, note it meets your requirements.

However, $D$ can move around anywhere on that circumscribed circle, so the circle going through $D$, as shown, would have different radii depending on where that point is, such as point $E$ instead as shown above. Overall, the locus for valid positions for $D$ would form a semi-circle of radius $4$ with center $C$.

This question is very nice but you can determine the radius of the circle with only these informations. I give you two different situations which satisfie hypotesis but the two radius are different. Instead of $8$ I use $5$ (Using Pythagorean triple in computation)

Case 1: Take as quadrilater a square with edge's length equal to $\frac{5}{\sqrt{2}}$. Thanks to this fact you have that the two diagonal are equal and of the circle has radius equal to $5$.

Case 2: Take the quadrilater $ABCD$ with coordinates $A=(0,0)$, $B=(3,0)$, $C=(4,2)$, $D=(0,4)$. Thanks to Pythagorean theorem we have that $\overline{BD} = 5$ as we want. Moreover, the ipotetic radius $AC$ has length equal to $\sqrt{4^2 + 2^2} = 2\sqrt{5}\neq 5$. So, if we show that $\overline{CD}$ is orthogonal to $\overline{BC}$ we are done. The angular coefficient of the line passing through $\overline{CD}$ is $-\frac{1}{2}$ and the angular coefficient of the line passing through $\overline{BC}$ is $2$; so $\overline{CD}$ is perpendicular to $\overline{BC}$.