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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! enter image description here

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    It depends on the distance between 2 centres. For instance, if the distance between centres is the sum of the radii of both circles, then they touch at 1 point only. – 2017-01-08
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    Do you want an algebraic proof or a compass and straight-edge proof? – 2017-01-08
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    @B.Goddard compass and straight-edge proof – 2017-01-08
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    Two points *at most*. You can have two circles that are completely disjoint, or two circles that are tangent (and intersect at only one point). – 2017-01-08

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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.

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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.

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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 $s$.