Tangents

Tangent from a point to a circle

Point $p$ , circle $(c, r)$ , $d = ∣ p - c ∣$ .

$d < r$ → no tangent ( $p$ is inside);
$d = r$ → one (at $p$ itself);
$d > r$ → two, and the tangent length is $d^{2} - r^{2}$ .

The tangent points lie on the circle centred at $c$ and on the circle with diameter $p c$ — so they are a circle-circle intersection:

vector<PD> tangentPoints(PD p, PD c, double r) {
    double d2 = norm2(p - c);
    if (d2 < r * r - EPS) return {};
    PD u = p - c;
    double a = r * r / d2;
    double h = r * sqrt(max(0.0, d2 - r * r)) / d2;
    PD base = c + u * a;
    PD perpU{-u.y * h, u.x * h};
    if (h < EPS) return {base};
    return {base - perpU, base + perpU};
}

The tangent length $d^{2} - r^{2}$ is exactly $pow (p, C)$ , the square root of the power of the point — which is why the radical axis (equal power) is also the locus of equal tangent length.

Common tangents to two circles

Configuration	External	Internal	Total
Separate ( $d > r_{1} + r_{2}$ )	2	2	4
Externally tangent	2	1	3
Intersecting	2	0	2
Internally tangent	1	0	1
One inside the other	0	0	0
Identical	—	—	infinite

External tangents keep both circles on the same side; internal tangents separate them.

Construction

Both families follow from one idea: a tangent line at signed distance $r_{1}$ from $c_{1}$ and $\pm r_{2}$ from $c_{2}$ . Writing the line as $a x + b y = c$ with $a^{2} + b^{2} = 1$ :

$a c_{1 x} + b c_{1 y} - c = r_{1}, a c_{2 x} + b c_{2 y} - c = \pm r_{2},$

which is a small linear system with two solutions per sign choice. Handling $r_{1} = r_{2}$ (parallel external tangents) as a separate case avoids a division by zero.

Tangent to a convex polygon from an external point

The two tangent vertices are where the polygon “turns away” from $p$ . With a CCW hull, each is found by binary search in $O (lo g n)$ : a vertex $v_{i}$ is the left tangent point iff both neighbours lie on the same side of line $p v_{i}$ .

// sign-based ternary/binary search on the hull
bool isTangent(P p, const vector<P>& h, int i, int sign) {
    int n = h.size();
    int a = orient(p, h[i], h[(i + 1) % n]);
    int b = orient(p, h[i], h[(i + n - 1) % n]);
    return a * sign >= 0 && b * sign >= 0;
}

Used for:

the visible portion of a polygon from a viewpoint,
adding a point to a hull in $O (lo g n)$ ,
Chan’s algorithm, whose inner loop is exactly this binary search.

Where tangents appear

Problem	Use
Shortest path around circular obstacles	the path is made of tangent segments and arcs
Belt / pulley length around circles	external tangents plus arcs
Visibility from a point	tangent lines bound the visible arc
Convex hull merging	tangent lines between two hulls
Rotating calipers	parallel tangent lines
Apollonius problems (circle tangent to three circles)	reduce with inversion
Light and shadow	tangent lines from a light source

The belt problem

The length of a taut belt around two pulleys of radii $r_{1}, r_{2}$ with centre distance $d$ :

crossed (figure-eight): $2 d^{2} - (r_{1} + r_{2})^{2} + (r_{1} + r_{2}) (π + 2 arcsin \frac{r _{1} + r _{2}}{d})$ ;
open: $2 d^{2} - (r_{1} - r_{2})^{2} + r_{1} (π + 2 α) + r_{2} (π - 2 α)$ with $α = arcsin \frac{r _{1} - r _{2}}{d}$ .

Deriving these from the tangent length is a good check that you have the configuration right.

whoashish's cp notes

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