Möbius transform which fixes real/imaginary axes must also transform $|z| = R_1$ to $|z| = R_2$ for $R_1,R_2 \in \mathbb{R}$

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Suppose there exists $f(z) = \frac{az+b}{cz+d}, \quad a,b,c,d \in \mathbb{C}$ such that if $z_0=K \in \mathbb{R}$ then $f(z_0)\in \mathbb{R}$ AND if $z_0 = L \in i\mathbb{R}$ then $f(z_0) \in i\mathbb{R}$

Prove that such a Möbius transform also maps circles centred at the origin to another circle centred at the origin (not necessarily of the same radius).

I have tried to tackle this by working backwards, showing that a Möbius transform which maps a circle of radius $R_1$ to a circle of radius $R_2$, will also for some $a,b,c,d$ also fix the imaginary and real axes, but I am not getting anywhere.