Your scenario looks as follows:

The center of the circle can move along the line (perpendicular bisector) which is orthogonal to the connection between $(1;2)$ and $(10;16)$.
The general equation of a circle is:
$$(X - M_x)^2 + (Y - M_y)^2 = r^2$$
Inserting your two points:
$$(1 - M_x)^2 + (2 - M_y)^2 = r^2$$
$$(10 - M_x)^2 + (16 - M_y)^2 = r^2$$
or
$$\begin{align}1 - 2M_x + M_x^2 + 4 - 4M_y + M_y^2 & = r^2\\
100 - 20M_x + M_x^2 + 256 - 32M_y + M_y^2 & = r^2\end{align}$$
Subtracting the second equation from the first:
$$-99 + 18M_x - 252 +28M_y = 0$$
Hence, we get linear expression for $M_y$ you could use in Excel:
$$M_y = - \frac{18}{28}M_x + \frac{351}{28}$$
Squared radius $r^2$ can be determined from the circle equation:
$$r^2 = (1 - M_x)^2 + (1 - M_y)^2$$
For $r = 10, 15, 18, 200$ the squared radius would be $100, 225, 324, 4000$.
$$\begin{aligned}
r^2 & = (1 - M_x)^2 + (- \frac{323}{28} + \frac{18}{28}M_x)^2 \\
& = 1 - 2M_x + M_x^2 + \frac{104329}{784} - \frac{11808}{784}M_x + \frac{324}{784}M_x^2\\
& = \frac{324}{784}M_x^2 - \frac{11024}{784}M_x + \frac{105113}{784}
\end{aligned}$$
This allows us to derive $M_x$ from $r^2$. Note that two solutions exist as the center of the circle can bei either on the right or on the left of $(9.5; 9)$.
Asking WolframAlpha reveals another formula suitable for Excel:
$$M_x = \frac{14\sqrt{1108r^2-93025} + 3299}{554}$$
To finally compute the $Y_i$ values corresponding to your $X_i$ values:
$$Y_i = M_y \pm \sqrt{r^2 - (X_i - M_x)^2}$$