= Solution
Write $r_a=\mu_{0,a}^2$. The stationary equations for
$$
V=\frac12r_1\phi_1^2+\frac12r_2\phi_2^2+g_0(\phi_1^2+\phi_2^2)^2
$$
are
$$
\phi_a\left[r_a+4g_0(\phi_1^2+\phi_2^2)\right]=0.
$$
For $r_1,r_2>0$, the disordered minimum is $(0,0)$. If $r_1<0$ and $r_1<r_2$, the first component orders with $\phi_1^2=-r_1/(4g_0)$ and $\phi_2=0$. If $r_2<0$ and $r_2<r_1$, the second component orders analogously.
The positive $r_2$ half of $r_1=0$ and the positive $r_1$ half of $r_2=0$ are continuous-transition lines. Along the negative diagonal $r_1=r_2<0$, the potential has an enhanced $O(2)$ symmetry and a circle of minima. Crossing that diagonal exchanges the two ordered axes and makes derivatives of the minimum free energy jump, so it is a <first-order phase transition> line. The origin, where the two continuous lines meet the first-order line, is a <bicritical point>.
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