We consider a $q$-dimensional Brownian motion $(W_r)_r.$ Let $x \in \mathbb{R}^q$ and $\delta \geq 0.$ Let $T:=\inf\{r \geq 0, W_r \notin B(x,\delta)\}.$
Prove that $P_x(T<\infty)=1,$ and that $W_T,$ under $P_x,$ is uniformly distributed on a sphere and determine it.
For the first part we know that since $\limsup_{r \to \infty}\frac{1}{\sqrt{r}}\Vert W_r \Vert_2\geq\limsup_{r \to \infty}\frac{1}{\sqrt{r}}|W^1_r|=\infty \ P_x$-a.s. where $\Vert. \Vert_2$ denotes the Euclidean norm that $\limsup_{r \to \infty}\Vert W_r \Vert_2=\infty \ P_x$-a.s. and therefore $T$ is finite a.s.
How can we prove that under $P_x, W_T$ is uniformly distributed on a sphere?
By translation, we may assume the starting point $x$ is $0$. Next, observe that the distribution of $W_T$ is rotation invariant, i.e, if $A$ is a Borel set on the sphere $\partial B(0,\delta)$ and $M$ is a rotation matrix, then $P(W_T \in A)=P(W_T \in M(A))$, because the Gaussian density is rotation invariant. Finally, use the classical fact that there is only one rotation invariant Borel probability measure on the sphere.