= Solution
Let $E_+$ and $E_-$ denote positive and negative real part at the circular exit. On $\{S<T(r)\}$, reflect the portion of the <planar Brownian motion> after $S$ by $z\mapsto-\overline z$. This reflection fixes the imaginary axis and preserves distances from the origin. The <Strong Markov property> and reflection symmetry show that the resulting path has the same law, its circular exit time is unchanged, and $E_+$ is exchanged with $E_-$. Therefore
$$
\mathbb P_1(S<T(r),E_+)=\mathbb P_1(S<T(r),E_-).
$$
An exit with negative real part must first cross the imaginary axis. An exit before $S$ has positive real part. The two points $ir,-ir$ have zero circular exit probability, since circular <harmonic measure> has no atoms. Hence
$$
\begin{aligned}
\mathbb P_1(E_+)&=\mathbb P_1(T(r)<S)+\mathbb P_1(S<T(r),E_+),\\
\mathbb P_1(E_-)&=\mathbb P_1(S<T(r),E_-).
\end{aligned}
$$
Subtracting proves
$$
\boxed{\mathbb P_1(T(r)<S)=\mathbb P_1(E_+)-\mathbb P_1(E_-).}
$$
This is the <reflection identity for Brownian exit from a half-disc>.
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