Intersection of two circles

The intersection of two circles is a point lying on both circles at once. Here we look at how $x^2 + y^2 = 25$ , centred at the origin with radius $5$ , meets $(x - 8)^2 + y^2 = 25$ , centred at $(8, 0)$ with the same radius.

Subtracting kills the quadratic terms

At an intersection both equations hold at once, so we may subtract one from the other.

(x^2 + y^2) - \left((x - 8)^2 + y^2\right) = 25 - 25

Expanding gives $16x - 64 = 0$ , that is $x = 4$ . Substituting into the first circle gives $y^2 = 9$ , so the intersection points are $(4, 3)$ and $(4, -3)$ .

The common chord

The linear equation that came out of the subtraction is the line through the two intersection points. To find where two circles meet, the standard method is to subtract first to get this line, then substitute it into a circle equation. The subtraction turns the problem into one we can already solve: the intersection of a circle and a line.

Because the two radii are equal in this example, the intersections lie on the perpendicular bisector of the segment joining the centres, and indeed $x = 4$ is the vertical line halfway between them. The common chord has length $6$ . In every case the two intersection points are symmetric about the line through the centres.

Counting before solving

How the circles meet is decided by the distance $d$ between centres compared with the radii.

Condition	Relationship
$d > r_1 + r_2$	separate, no intersection
$d = r_1 + r_2$	externally tangent
$\|r_1 - r_2\| < d < r_1 + r_2$	crossing at two points
$d = \|r_1 - r_2\|$	internally tangent
$d < \|r_1 - r_2\|$	one inside the other

Here $d = 8$ , $r_1 + r_2 = 10$ and $|r_1 - r_2| = 0$ , so the circles cross at two points, as we found. The count is settled by the centres and the radii alone, before any equation is solved.

The large dots on the graph are those two intersection points.

Circle, Wikipedia