Circle-Circle Intersection

Circles $(c_1,r_1)$ and $(c_2,r_2)$ with $d=|c_1-c_2|$.

The cases

Condition	Result
$d>r_1+r_2$	separate — 0 points
$d=r_1+r_2$	externally tangent — 1 point
$|r_1-r_2|<d<r_1+r_2$	2 points
$d=|r_1-r_2|$, $d>0$	internally tangent — 1 point
$d<|r_1-r_2|$	one inside the other — 0 points
$d=0$, $r_1=r_2$	identical — infinitely many

Test with squared distances to stay exact: compare $d^2$ against $(r_1+r_2{)}^2$ and $(r_1-r_2{)}^2$.

The construction

Both intersection points lie on the radical line, perpendicular to $c_1{c}_2$. Let

$a=\frac{d^2+r_1^2-r_2^2}{2d},h=\sqrt{r_1^2-a^2}.$

Then the base point is $p=c_1+a\hat{u}$ with $\hat{u}=(c_2-c_1)/d$, and the intersections are $p\pm h{\hat{u}}^{\perp }$.

vector<PD> circleCircle(PD c1, double r1, PD c2, double r2) {
    PD u = c2 - c1;
    double d2 = norm2(u), d = sqrt(d2);
    if (d < EPS) return {};                                  // concentric
    if (d > r1 + r2 + EPS) return {};
    if (d < fabs(r1 - r2) - EPS) return {};
 
    double a = (d2 + r1*r1 - r2*r2) / (2 * d);
    double h2 = r1*r1 - a*a;
    PD base = c1 + u * (a / d);
    if (h2 < EPS) return {base};                             // tangent
    double h = sqrt(h2);
    PD perpU{-u.y / d * h, u.x / d * h};
    return {base - perpU, base + perpU};
}

The radical line and radical centre

The radical axis of two circles is the locus of points with equal power to both:

$\mathrm{pow}(p,C)=|p-c{|}^2-r^2.$

Setting the two powers equal gives a line (the $x^2+y^2$ terms cancel):

$2(c_2-c_1)\cdot p=|c_2{|}^2-|c_1{|}^2-r_2^2+r_1^2.$

For three circles the three radical axes meet at the radical centre — the point from which tangent lengths to all three are equal. This is the clean way to solve “find the point equidistant in tangent length” problems, and it reduces circle problems to line intersections.

Areas

Lens (intersection) area of two overlapping circles:

$A=r_1^2\left({\theta }_1-\sin {\theta }_1\right)/2+r_2^2\left({\theta }_2-\sin {\theta }_2\right)/2,$

with ${\theta }_1=2\arccos \frac{d^2+r_1^2-r_2^2}{2d{r}_1}$ and symmetrically for ${\theta }_2$.

Union of $n$ circles: for each circle, walk its boundary and determine which arcs are not covered by any other circle; then apply a Green’s theorem style sum over arcs and chords. $O(n^2\log n)$.

Related constructions

Task	Method
Circle through 3 points	perpendicular bisectors, or the determinant formula
Circle tangent to 2 circles with radius $r$	intersect circles of radii $r_1\pm r$ and $r_2\pm r$
Apollonius circle	the locus with $|PA|/|PB|=k$
Common tangent lines	see the tangents page
Circle inversion	maps circles and lines to circles and lines
Smallest enclosing circle	Welzl

Precision

Near-tangency

When $d$ is close to $r_1+r_2$, $h^2$ is a tiny difference of large numbers and loses most of its significant digits. Always clamp with max(0.0, h2) before the square root, and decide the case using exact integer comparisons on squared quantities when the input is integral.

See also: Circle-Line Intersection · Tangents · Minimum Enclosing Circle