We use complex numbers to prove that there are at most $4$ such points unless both transmitter and receiver are at the center.
Identify the circular room with the unit circle $|z|=1$ in the complex plane, and let $r$ and $t$ be the complex numbers corresponding to the receiver and transmitter, with $|r|<1$ and $|t|<1$. If $z$ is a point of reflection then the condition $\theta_r = \theta_t$ comes down to $(z-r)(z-t)$ being a real multiple of $z^2$; that is, to the ratio $(z-r)(z-t)/z^2$ being a real number. Write $$ (z-r)(z-t) / z^2 = 1 - (r+t) z^{-1} + rt z^{-2}, $$ and note that a complex number $w$ is real if and only if it equals its own complex conjugate $\overline w$. Since $z$ is on the unit circle, $\overline z = z^{-1}$, so our condition is $$ \overline{rt} z^2 - (\overline r + \overline t) z + (r+t) z^{-1} - rt z^{-2}. $$ Multiplying by $z^2$ yields a polynomial of degree $4$ in $z$. Thus there are at most $4$ solutions, even without the condition $|z|=1$, unless the polynomial vanishes identically, in which case every $z$ is a solution. But the polynomial vanishes identically if and only if $r+t = rt = 0$, which is to say $r=t=0$, so we recover the degenerate case where receiver and transmitter are both in the center of the circular room.