Answer: 272

Small Hint:

If each $22$-chord subtends central angle $\alpha$ and each $20$-chord subtends $\beta,$ then $4\alpha + 2\beta = 360^\circ,$ so $\sin\frac{\beta}{2} = \cos\alpha$

Big Hint:

The law of cosines gives $\cos\alpha = 1 - \frac{242}{r^2},$ while $\sin\frac{\beta}{2} = \frac{10}{r};$ equating them yields a quadratic in $r$

Solution:

Half a $20$-chord gives $\sin\frac{\beta}{2} = \frac{10}{r},$ and the law of cosines on the isosceles triangle with legs $r$ and base $22$ gives $22^2 = 2r^2(1 - \cos\alpha),$ so $\cos\alpha = 1 - \frac{242}{r^2}.$ Equating, $$\begin{aligned} 1 - \frac{242}{r^2} &= \frac{10}{r} \\ &\quad\Longrightarrow\quad r^2 - 10r - 242 \\ &= 0, \end{aligned}$$ so $r = 5 + \sqrt{267}$ (taking the positive root).

Therefore $p + q = 5 + 267 = 272.$