
how to prove this?
Let two intersecting circles be represented by equations x^2 + y^2 + Ax + By + C =0 and x^2 + y^2 + ax + by + c = 0.
For any number k that is not equal to 1, show that the equation of the circle through the intersection points of the two circles is x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0.
Conversely, also show that every such circle may be represented by such an equation with a suitable k.

Re: how to prove this?
First; every circle can be written in the form $x^2+y^2+ux+vy+w=0$ and every such equation represents a circle. Thus $x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0$ is a circle iff $k\ne 1$.
Also for the intersection point $(x,y)$ we have $x^2 + y^2 + Ax + By + C=0$ and $x^2 + y^2 + ax + by + c=0$ hence for such a point $x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0$ so the intersection points of the two circles lie on the third circle.

Re: how to prove this?
Your statement of the problem is not quite correct so I restated it. The previous response is not quite correct; here is a complete solution.
http://i61.tinypic.com/9hksnl.png