I have been looking at the Schwarzschild metric presented to me as the following within lectures:

$$ds^2=-\frac{\textrm{d}r^2}{1+\frac{\gamma}{r}}-r^2\textrm{d}\theta^2-r^2\sin^2\theta\textrm{d}\phi^2+c^2\left(1+\frac{\gamma}{r}\right)\textrm{d}t^2,$$

where $\gamma=-\frac{2GM}{c^2}$.

For circular motion the radius can be taken as constant and $\theta=\frac{\pi}{2}$ can be set without a loss of generality.

From this metric and taking into consideration the constant variables,the Lagrangian is

$$L= -\frac{\dot{r}^2}{1+\gamma/r}-r^2\dot{\phi}^2+c^2\left(1+\gamma/r\right)\dot{t}^2.$$

The Euler-Lagrange equations are

$$\frac{\textrm{d}}{\textrm{d}s}\left[\frac{\partial L}{\partial \dot{x}^{\mu}}\right]-\frac{\partial L}{\partial x^{\mu}}=0.$$

My question arises here, up until this point whenever a variable has been given as a constant, in the corresponding part of the metric, I have been able to effectively eliminate this part of the metric as the derivative of a constant is equal to 0. From inspecting the Lagrangian I can see that this has been done in the case of $\theta$, but even though r has been stated to be a constant it is kept in the statement of the Lagrangian, why is this?

Through further manipulations considering the case of the Lagrangian for $\mu=1$, it is possible to get the following

$$r\dot{\phi}^2=\frac{GM}{r^2}\dot{t}^2,$$

whereby a statement of Kepler's law is obtained

$$\Omega^2=\left(\frac{\textrm{d}\phi}{\textrm{d}t}\right)^2=\frac{GM}{r^3}.$$

I am having trouble understanding how this is valid given that it seems to rely on $\dot{r}$ being kept in the Lagrangian even though $r$ is stated as a constant and therefore $\dot{r}$ should be 0.