Solution Let X be the tangency point of L_1 and AB. Let CX meets L_1 in Y_1 and CX meets L_2 in Y_2. We shall establish that Y_1 = Y_2.
Let C' be the diameter of L_2 and let CC' meets AB (or its extension) at D. Because CP = CQ then it's easy to see that CD \perp PQ. Let s = CM = CN = CP = CQ.
By power point on L_1: CX.CY_1 = CM^2 = s^2.
By similarity of \triangle CXD \sim \triangle CC'Y_2 and \triangle CC'P \sim \triangle CDP : CX.CY_2 = CD.CC' = CP^2 = s^2
Thus CY_1 = CY_2 which means Y_1 = Y_2. Now we know that L_1 meets L_2 in at least one point Y = Y_1 = Y_2. Suppose they meet in another point Z \neq Y
Let CZ meets L_1 in CZ and X_1 \neq X. By power point we again have:CX_1.CZ = s^2.
Let CZ meets AB in X_2. Similar as above, because Z \in L_2, we can repeat the similarity argument to establish: CX_2.CZ = s^2. This means that CX_1 = CX_2, thus X_1 = X_2 \in L_1, AB but that is a contradiction because L_1 is only tangent to AB, and can only meet AB at X.
Now that we've established the tangency between L_1 and L_2, it's clear that C,X,Y_1/Y_2 all lie on one line from the proof.
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