How to prove the the general equation of the family of circles through the intersection points of two circles? Let $ \mathscr{C_{1}} $ and $ \mathscr C_{2} $ be two intersecting circles determined by the equations $$ x^2 + y^2 + A_{1}x + B_{1}y + C_{1} = 0 $$ and $$ x^2 + y^2 + A_{2}x + B_{2}y + C_{2} = 0. $$ For any number $ k≠−1 $, show that $$ x^2 + y^2 + A_{1}x + B_{1}y + C_{1} + k ( x^2 + y^2 + A_{2}x + B_{2}y + C_{2} ) = 0 $$ is the equation of a circle through the intersection points of $\mathscr{C_{1}}$ and $\mathscr C_{2}$. Show, conversely, that every such circle may be represented by such an equation for a suitable $ k $.



I'm stuck in this question, developing the given equation I could show that it assumes the form of 
$$ x^2 + y^2 + \frac{A_{1}+kA_{2}}{1+k}x + \frac{B_{1}+kB_{2}}{1+k}y + \frac{C_{1}+kC_{2}}{1+k} = 0 $$ 
which  proves that for $ k≠−1 $ the given equation is the equation of a circle. But I'm not entirely sure if it really proves that it is the equation of the family of circles through the intersection points of $ \mathscr{C_{1}} $ and $ \mathscr C_{2} $. I have no clue about how to prove the converse statement.

Let the two points of the intersection be A and B and consider the straight line AB then, since the equation  $x^2 + y^2 + \frac{A_{1}+kA_{2}}{1+k}x + \frac{B_{1}+kB_{2}}{1+k}y + \frac{C_{1}+kC_{2}}{1+k} = 0$ is the equation of a circle of centre $(\frac{A_{1}+kA_{2}}{2(1+k)}, \frac{B_{1}+kB_{2}}{2(1+k)})$ pasing through the intersection points, you have to prove that the set of all those points for $k \in \mathbb{R} - \left\{1\right\}$ is a straight line except for a point (because since all the circles pass through the points, then it should be the bisection). Well, actually, we are forgetting the case $k = -1$ but check where is the centre.
