how to prove this?

• Aug 21st 2014, 05:46 PM
enochxx7
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.
• Aug 23rd 2014, 11:56 PM
zzephod
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.
• Aug 24th 2014, 05:59 PM
johng
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