Sketch of solution, using points given in image. We claim that $MX$ is tangent to both circles. To show that $MX$ is tangent to ${EXF}$, suffices to show $MX^2=ME\times MF$, but $ML^2=ME\times MF$ because $(FELK)$ is harmonic. To show $MX$ is tangent to the incircle, by Pascal, the pairs of tangency points meet at $L$ and $K$. Then, we can eliminate points $ABCDEF$. The rest can be done using linearity of PoP to the two circles, because the sum of powers from $L$ and $K$ to the incircle is simply $LK^2$. (Proof by Miguel point). I hope to see an easier solution :>

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Let $A_1$ be the tangency point of $\gamma$ to $AB,$ and define $B_1,C_1,D_1$ cyclically. It suffices to show that $(EXF)$ and $\gamma$ are both orthogonal wrt $(KL).$ It is a well known configuration that $(EF, LK)$ is harmonic and inversion wrt $(KL)$ swaps $E,F.$ So $(EXF)$ is done. By Pascal's on $A_1A_1B_1D_1D_1C_1$ we know that $AC$ is concurrent with $A_1B_1,C_1D_1$ (since $A_1C_1,B_1D_1,AC$ are concurrent). By Pascal's on $A_1A_1B_1C_1C_1D_1$ and $A_1B_1B_1C_1D_1D_1$ we see that the concurrency point lies on $EF$ so it is $K.$ So $K$ lies on $A_1B_1,C_1D_1$ which are the polars of $B,D$ under inversion wrt $\gamma.$ So $BD$ is the polar of $K.$ Similarly, $AC$ is the polar of $L.$ But $K$ lies on $AC$ and $L$ lies on $BD.$ So $(KL)$ maps to itself under inversion wrt $\gamma$ as desired. $\blacksquare$

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Another way: It's easy to see that $(KL,FE) = -1$, so $(KXL)$ is orthogonal with every circle go through $E,F$. Hence, $(KXL)$ is orthogonal to $(EXF)$. Then, invert wrt $\gamma$ to get that $(KXL)$ is orthogonal to $\gamma$