Rather than embarking on some pretty involved computations of conditional distributions, one should rely on one of the main assets of Gaussian families, namely, the...
Key feature: In Gaussian families, conditioning acts as a linear projection.
Hence, as the OP suggested, one could do worse than to start from a representation of $(X,Y)$ by standard i.i.d. Gaussian random variables $U$ and $V$, for example,
$$
X=\mu_x+\sigma_xU\qquad
Y=\mu_y+\sigma_y(\rho U+\tau V)$$ where the parameter $\tau$ is $$\tau=\sqrt{1-\rho^2}
$$
Since $\sigma_x\ne0$, the sigma-algebra generated by $X$ is also the sigma-algebra generated by $U$ hence conditioning by $X$ or by $U$ is the same. Furthermore, constants and functions of $X$ or $U$ are all $U$-measurable while functions of $V$ are independent on $U$, thus,
$$
\mathrm E(Y\mid X)=\mu_y+\sigma_y(\rho U+\tau \mathrm E(V))=\mu_y+\sigma_y \rho U
$$
which is equivalent to
$$
\color{red}{\mathrm E(Y\mid X)=\mu_y+\rho\frac{\sigma_y}{\sigma_x}(X-\mu_x)}
$$
Likewise, when computing conditional variances conditionally on $X$, deterministic functions of $X$ or $U$ should be considered as constants, hence their conditional variance is zero, and functions of $V$ are independent on $X$, hence their conditional variance is their variance. Thus,
$$
\mbox{Var}(Y\mid X)=\mbox{Var}(\sigma_y\tau V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V)
$$
that is,
$$
\color{red}{\mbox{Var}(Y\mid X)=\sigma_y^2(1-\rho^2)}
$$
Finally, the event $$A=[X>\mu_x,Y>\mu_y]$$ is also
$$
A=[U>0,\rho U+\tau V>0].
$$
To evaluate $\mathrm P(A)$, one can turn to the planar representation of couples of independent standard Gaussian random variables, which says in particular that the distribution of $(U,V)$ is invariant by rotations. The event $A$ means that the direction of the vector $(U,V)$ is between the angle $\vartheta$ in $(-\pi/2,\pi/2)$ such that $$\tan(\vartheta)=-\rho/\tau$$ and the angle $\pi/2$. Thus,
$$\mathrm P(A)=\frac{\pi/2-\vartheta}{2\pi}$$
that is,
$$
\color{red}{\mathrm P(X>\mu_x,Y>\mu_y)=\frac14+\frac1{2\pi}\arcsin\rho}
$$
Numerical application: If $\mu_x=2$, $\mu_y=-1$, $\sigma_x=2$, $\sigma_y=1$ and $\rho=-\sqrt3/2$, then
$$
\mathrm E(Y\mid X)=-1+\sqrt3/2-(\sqrt3/4)X\qquad
\mbox{Var}(Y\mid X)=1/4
$$
and $\tau=1/2$, hence $\vartheta=\pi/3$ and $$\mathrm P(A)=1/12$$
I don't have a combinatorial meaning, but you can think of it as follows.
$(X,Y)$ is the result of applying an affine transformation to a pair
$(W,Z)$ of independent standard normal random variables. Many such
transformations exist, and one in particular is
$$\begin{align*}
X &= \mu_x + \sigma_x W\\
Y &= \mu_y + \rho \sigma_y W + \sqrt{1-\rho^2} \sigma_y Z
\end{align*}$$
See for example
this set of slides. The contours of the joint density (points at
equal height above the $x$-$y$ plane) are ellipses centered at $(\mu_x,\mu_y)$.
Best Answer
Using
$$f(X|X+Y=5)=\frac{f(X+Y=5|X)f(X)}{f(X+Y=5)}$$
The pieces are
$$\begin{split}Y|X&\sim\text{Normal}\left( \mu_2+\rho\sigma_2\left(\frac{x-\mu_1}{\sigma_1}\right), (1-\rho^2)\sigma_2^2\right)\\ X&\sim\text{Normal}\left(\mu_1, \sigma_1^2\right)\\ X+Y&\sim\text{Normal}\left(\mu_1+\mu_2, \sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2\right)\end{split}$$
The conditional density is $$f(X|X+Y=5)=\frac{\sqrt{\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2}}{\sqrt{2\pi}(1-\rho^2)^{1/2}\sigma_1\sigma_2}\exp-\frac 1 2\left\{\frac{(5-x-\mu_2-\rho\sigma_2\left(\frac{x-\mu_1}{\sigma_1}\right)^2}{(1-\rho^2)\sigma_2^2}+\frac{(x-\mu_1)^2}{\sigma_1^2}-\frac{(5-\mu_1-\mu_2)^2}{\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2}\right\}$$
Do a lot of algebra and even then focusing only on the terms involving $x$ to get a normal pdf with mean:
$$E(X|X+Y=5)=\frac{\left(5-\mu_2+\frac{\rho\sigma_2\mu_1}{\sigma_1}\right)\left(1+\frac{\rho\sigma_2}{\sigma_1}\right)\sigma_1^2+\mu_1(1-\rho^2)\sigma_2^2}{\left(1+\frac{\rho\sigma_2}{\sigma_1}\right)^2\sigma_1^2+(1-\rho^2)\sigma_2^2}$$
A brief simulation confirms the result for $\mu_1=2,\mu_2=2,\sigma_1=2,\sigma_2=2,\rho=.5$, we have an expectation of
2.5and a simulation gives2.51738.