Please help...?

Question

Prove that two different circles cannot intersect each other at more than two points. The whole, correct proof, please!!! I need it urgently

Answer 1

Sorry. I'll just give you a big hint.

It suffices to show that given 3 non-collinear points, there's at most 1 circle they lie on. (Actually, they lie on exactly one circle.)

That follows immediately if you can prove there's only one possible point that could be the center of the circle (because if you have the center, the radius can be instantly calculated; it's just the distance from the center to any of the three points).

So you're trying to prove given three points R, S, and T, there's at most one point C equidistant from them.

Unfortunately, you haven't said what you're studying. Euclidean Geometry? Precalculus? The techniques available to you depend a lot on that, and hence the appropriate method of proof may be different.

In particular -- do you have graphing and Cartesian coordinates available?

Answer 2

Hmmm ! Good question !

How about something along these lines;

h(x, y) = x² + y² + 2gx + 2fy + c = 0 is one circle
H(x, y) = x² + y² + 2Gx + 2Fy + C = 0 is the other.

Any points which simultaneously lie on both circles must satisfy :
α.h(x, y) + β.H(x, y) = 0 . . .in particular, the case where α and β are chosen to eliminate the x² and y² terms.

In my case say, α = 1 and β = -1.

The resulting equation is (2g - 2G)x + (2f - 2F)y + (c - C) = 0

All ponts common to the two circles must lie along this line, so use it to eliminate x (or y) in one of the original equations of your circles. A quadratic results. Maximum number of solutions = 2.

Sorry I couldn't think up a more elegant "solution" . . hope it will do.

Answer 3

x^2+y^2+ax+by+c=0 (1)
x^2+y^2+px+qy+s=0 (2)
You must solve this system
1) substract one from the other

(a-p)x+(b-q)y +c-s =0 (3)
Now solve the system formed by (1) and (3)
For this calculate y from (3)

y= [s-c-(a-p)x]/(b-q) (4)

This expression put into equation (1) gives you a second degree
equation in x .This one can have at most two solutions which are the x numbers of the intercepts.After this you get the y's from
(4)
So there are at most two points of interception