How should we prove that two circles can intersect at two points( at least)?

Question

Assume that there are two distinct circles with centres C and D respectively. I feel that these two circles can intersect at two points but I don't know how to prove that they can intersect at two points! However I tried to prove it by construction like this- "I firstly construct a circle and then I again construct other circle with a compass such that they both intersect each other at two points." Is my way of proving correct? If not,then please provide an appropriate proof for this? THANKS!

Answer 1

The key point to a proof is that if you have three non-collinear points, they determine a unique circle. (So two distinct circles can intersect in at most two points.) You can prove this by construction: The center of the circle will be the intersection of the perpendicular bisectors of the segments joining pairs of the points.

Answer 2

Of course, there's the analytic approach.

Moving one of the circles to the origin, the equations are

$x^2+y^2 = r^2$ and $(x-a)^2+(y-b)^2 = s^2$.

Subtracting the second from the first,

$2ax-a^2+2by-b^2 = r^2-s^2$ or $2ax+2by = r^2-s^2+a^2+b^2 =c$.

If $b=0$, then $a \ne 0$ (or else the two circles are concentric and then have zero or an infinite number of intersections) so that $x = c/2$ and $y^2 =r^2-x^2 =r^2-\frac{c^2}{4} $ which has zero, one, or two solutions depending on the sign of $4r^2-c^2$.

If $b\ne 0$, then $y = \frac{c}{2b}-\frac{ax}{b}$. putting this in the first equation,

$\begin{array}\\ r^2 &=x^2+\left(\frac{c}{2b}-\frac{ax}{b}\right)^2\\ &=x^2+\frac{c^2}{4b^2}-\frac{acx}{b^2}+\frac{a^2x^2}{b^2}\\ &=x^2\frac{a^2+b^2}{b^2}-x\frac{ac}{b^2}+\frac{c^2}{4b^2}\\ \text{or}\\ 0 &=x^2\frac{a^2+b^2}{b^2}-x\frac{ac}{b^2}+\frac{c^2}{4b^2}-r^2\\ &=x^2\frac{a^2+b^2}{b^2}-x\frac{ac}{b^2}+\frac{c^2-4b^2r^2}{4b^2}\\ \end{array} $

This is a quadratic in $x$ and so has zero, one, or two real roots depending on the sign of the discriminant which is

$\begin{array}\\ \frac{a^2c^2}{b^4}-\frac{(a^2+b^2)(c^2-4b^2r^2)}{b^4} &=\frac{a^2c^2-(a^2+b^2)c^2+4b^2r^2(a^2+b^2)}{b^4}\\ &=\frac{b^2(4r^2(a^2+b^2)-c^2)}{b^4}\\ &=\frac{4r^2(a^2+b^2)-c^2}{b^2}\\ \end{array} $

You can substitute the actual expression for $c$ to get results involving only $a, b, r,$ and $s$.

Answer 3

I think I would draw the line through $C$ and $D$. Then draw the circle centered at $D$, or at least an arc of it that cuts the segment $CD$ at $P$. Set the compass for the radius of the circle centered at $C$, but draw the circle of that radius centered at $P$. If $C$ is inside that circle, then there are two intersection points.