Let the circle $ATX$ and the line $A_2B_1$ meet again at the point $Y$, and let the circle $BTX$ and the line $A_1B_2$ meet again at the point $Z$. Notice that the lines $AB, AY$ and $AZ$ are all parallel to the line $A_1B_1$ to deduce that the points $A, B, Y$ and $Z$ are collinear.

Next, consider the cyclic quadrangles $A_1B_1A_2B_2$, $A_1B_1B_2T$ and $BTB_2Z$ to get successively $\angle B_2A_2Y = \angle B_1A_1B_2 = \angle B_1TB_2 = \angle BZB_2$, and infer thereby that the points $A_2, B_2, Y$ and $Z$ are co-cyclic. It then follows that the lines $TX, A_1B_2$ and $A_2B_1$ are the radical lines of the pairs of circles $ATX$ and $BTX, BTX$ and $A_2B_2YZ$, and $A_2B_2YZ$ and $ATX$, respectively, whence the conclusion.

Apply an inversion of pole $T$ to rephrase the statement as follows
Let $\delta_1$ and $\delta_2$ be two parallel lines, let $T$ be a point in the complement of $\delta_1 \cup \delta_2$, and let $\ell_1$ and $\ell_2$ be two lines through $T$. The lines $\ell_1$ and $\ell_2$ meet $\delta_1$ at points $A$ and $B$, respectively, and $\delta_2$ at points $A_1$ and $B_1$, respectively. Let further $X$ be a point in the complement of $\delta_1 \cup \delta_2 \cup \ell_1 \cup \ell_2$. The lines $AX$ and $BX$ meet $\delta_2$ at points $A_2$ and $B_2$, respectively. Prove that the circles $TA_1B_2$ and $TA_2B_1$ meet again at a point on the line $TX$.

In other words, we must prove that the line $TX$ is the radical line of the circles $TA_1B_2$ and $TA_2B_1$. It is sufficient to show that the point $Y$ where the line $TX$ meets $\delta_2$ is of equal powers with respect to the two circles. To this end, apply Menelaus' theorem to triangles $AA_1A_2$ and $BB_1B_2$ and transversal $TXY$, to get, upon multiplication and suitable rearrangement of factors,
$$YA_1 \cdot YB_2 = (YA_2 \cdot YB_1) \left( \frac{TB}{TB_1} \cdot \frac{TA_1}{TA} \right) \left( \frac{XA}{XA_2} \cdot \frac{XB_2}{XB} \right).$$
Finally, notice that the last two products in the parentheses above both equal 1, by similarity of triangles $TAB$ and $TA_1B_1$, and $XAB$ and $XA_2B_2$, to conclude that $Y$ is indeed of equal powers with respect to the circles $TA_1B_2$ and $TA_2B_1$.