Prove that all the three lines are concurrent.

Question

There are $3$ circles centered at $A,B,C$ known as $\alpha,\beta,\gamma$ respectively. All of them intersecting each other at $2$ points.
Let $\alpha \cap \beta=C_1,C_2$ ; $\gamma \cap \beta=A_1,A_2$ ; $\alpha \cap \gamma=B_1,B_2$
Then prove that $A_1A_2,B_1B_2,C_1C_2$ are concurrent.

Answer 1

The three circles are$$\alpha:\space (x-a_1)^2+(y-a_2)^2=r_a^2\\\beta:\space (x-b_1)^2+(y-b_2)^2=r_b^2\\\gamma:\space (x-c_1)^2+(y-c_2)^2=r_c^2$$ By simple subtraction we get the lines $C_1C_2,B_1B_2$ and $A_1A_2$. We have $$C_1C_2:\space 2(b_1-a_1)x+2(b_2-a_2)y=r_a^2-r_b^2\\B_1B_2:\space 2(c_1-a_1)x+2(c_2-a_2)y=r_a^2-r_c^2\\A_1A_2:\space 2(c_1-b_1)x+2(c_2-b_2)y=r_b^2-r_c^2$$ To prove that the lines are concurrent we can solve two equations and verify that the solution satisfies the other equation. But it is much better to see that a linear combination of two of them gives the third one.

In fact, it is verify that $B_1B_2-A_1A_2=C_1C_2$:

$$B_1B_2-A_1A_2=2(c_1-a_a-c_1+b_1)x+2(c_2-a_2-c_2+b_2)y=r_a^2-r_c^2-r_b^2+r_c^2=C_1C_2$$