Let's assume a bivariate normal distribution with center $\mu$ and covariance matrix $\Sigma$. Let a circle $C$ be given as $C=\{x\in\mathbb{R}^2:||x-\mu||\leq R\}$. I would like to calculate the probability of being in the circle, i.e.
$$ P=\int_{C} f(x) dx $$
At least approximately.
The covariance matrix may be general, but an answer with diagonal covariance matrix will be appreciated, too.
Setup hint:
Start with finding a suitable matrix $A^{2\times2}$ such that $AA^T=\Sigma$. Then write $X=AU+\mu$ where $U=(U_1,U_2)^T$ is a random vector such that $U_1,U_2$ are iid and have standard normal distribution.
Then $X$ has the distribution that you mention and:$$\Pr\left(X\in C\right)=\Pr\left(\left(X-\mu\right)^{T}\left(X-\mu\right)\leq R^{2}\right)=\Pr\left(U^{T}A^{T}AU\leq R^{2}\right)$$
This is not a full answer, but it can make things easyer: the $U_i$ are iid and have a "nice" distribution without any annoying parameters.