Find $\mathbb{P} [ \| Z+a\| \le r \mid \|Z \| \le r ]$ where $Z$ is an i.i.d. Gaussian vector How to find 
\begin{align}
\mathbb{P} [ \| Z+a\| \le r \mid \|Z \| \le r ]
\end{align}
where $Z \in \mathbb{R}^n$ is an i.i.d. Gaussian vector?   Here $\| \cdot\|$ is the Euclidean norm. 
If the close form is difficult to get it would be interesting to see an upper bound that is asymptotically tight as $n\to \infty$.  
If $\| a\| \ge 2r $ then the probability is zero.  So we have to assume that  $\| a\| \ge 2r $. 
One can approach this question by using Bayes rule 
\begin{align}
\mathbb{P} [ \| Z+a\| \le r \mid \|Z \| \le r ]= \frac{ \mathbb{P} [ \| Z+a\| \le r , \|Z \| \le r ]}{ P[\|Z \| \le r ]},
\end{align}
Note that $P[\|Z \| \le r ]=1-\frac{\Gamma(n/2,r^2/2)}{\Gamma(n/2)}$.  
So, the question boils down to finding the joint probability $\mathbb{P} [ \| Z+a\| \le r , \|Z \| \le r ]$.
As suggested in sketch of a proof by  @kimchi lover   we should define
\begin{align}
(R,S)=(\|Z+a\|, \|Z\|)= \left(\sqrt{ (Z_1+A)^2 +Q}, \sqrt{ Z_1^2 +Q}   \right)
\end{align}
where $Q$ is a chi-square distribution.  In this case
\begin{align}
\mathbb{P} [ \| Z+a\| \le r , \|Z \| \le r ]=\mathbb{P} [R \le r , S \le r ]= \int_0^r \int_0^r f_{R,S}(x,y)dxdy.
\end{align}
However, it is not very clear to me how to find the joint distribution  $f_{R,S}(x,y)$.
 A: Assuming $\|v\|$ denotes the Euclidean length of $v\in\mathbb{R}^n$, you can use this method, involving a certain amount of calculus.  By the rotational invariance of mean-$0$ i.i.d. Gaussians,  you may as well assume $a$ is of the form $(A,0,\ldots,0)$, with $A\ge0$.  Now it is easy to express the joint distribution of the pair $(R,S) = (\|Z+a\|, \|Z\|)$ in terms of the independent r.v.s $Z_1$ and $\sum_{i>1} Z_i^2$. The former is $N(0,1)$ and the latter is chi-squared with $n-1$ degrees of freedom.  Your desired conditional probability is the ratio of two integrals, each over a 2-dimensional region. 
ADDED, edited 17 July 2017.
Writing $R=\sqrt{(Z_1+A)^2+Q}$ and $S=\sqrt{Z_1^2+Q}$, where $Z_1$ is $N(0,1)$ and $Q$ is $\chi_{n-1}^2$, the numerator integral (for $P(R\le r, S\le r)$) is $$E \,P( R\le r, S\le r | Q) = E \,\,I_{Q\le r^2} P ( (r^2-Q)^{1/2}-A \le Z_1 \le (r^2-Q)^{1/2} | \, Q).$$  (Here $I_{Q\le r^2}$ is the indicator r.v. for the event $[Q\le r^2].$)
If $f$ denotes the $\chi_{n-1}^2$ density, the numerator integral is
$$ \int_0^{r^2-A^2/4}f(q) \left( \Phi( (r^2-q)^{1/2} - \Phi((r^2-q)^{1/2}-A) \right) \,dq.$$
(The upper range of integration encodes the condition that the intersection of the events $[(Z_1+A)^2+Q\le r^2]$ and $[Z_1^2+Q\le r^2]$ is non-empty. Conditional on $Q$, each of these events bounds $Z_1$ to an interval: $-\Delta -A \le Z_1 \le \Delta -A$ and $-\Delta \le Z_1 \le \Delta$, respectively, where $\Delta=\sqrt{r^2-Q}$.  The intersection is nontrivial when $\Delta-A \ge -\Delta$.  Then the intersection is $[\Delta-A \le Z_1 \le \Delta].$)
The denominator integral is $$ \int_0^{r^2} f(q) \left( \Phi( (r^2-q)^{1/2} - \Phi(-(r^2-q)^{1/2}) \right) \,dq.$$
These integrals are formally 1-dimensional, but the use of the $\Phi$ function can be considered as sweeping 2-dimensional integral dirt under the rug.
