Strong-shock polar for a perfect gas ($u_Y^2=(U-u_X)(u_X-qU),q=(\gamma -1)/(\gamma +1)$)

The normal velocity is reduced by $q$ in the strong-shock limit while tangential velocity is unchanged. If the upstream speed is $U$ and the shock normal makes angle $\beta$ with its direction, then $u_X=U[1-(1-q){\cos }^2\beta ]$ and $u_Y=\pm U(1-q)\cos \beta \sin \beta$. Eliminating the angle gives the displayed circle, centered at $\gamma U/(\gamma +1)$ with radius $U/(\gamma +1)$. It is the limiting shock polar in downstream velocity space.

For the circle with center distance $a=\gamma U/(\gamma +1)$ and radius $R=U/(\gamma +1)$, the most deflected ray from the origin is tangent. The right triangle formed by the origin, center and tangency point gives $\sin {\delta }_{\max }=R/a=1/\gamma$. This is a velocity-vector deflection angle, not the angle of the shock normal.