# Thread: how to prove this?

1. ## 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.

2. ## 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.

3. ## 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.