Circle ($S^1$)

The circle is the subspace $S^1=\{z\in \mathbb{C}:|z|=1\}$, equivalently the quotient of a closed interval obtained by identifying its endpoints.

A Euclidean circle with center $c\in {\mathbb{R}}^2$ and radius $r>0$ is the set of points in the Euclidean plane at Euclidean distance exactly $r$ from $c$:It is the boundary curve, while the closed disk includes all points with $\Vert x-c\Vert \le r$. The center belongs to that closed disk but not to the Euclidean circle.

For example, center $c=(2,-1)$ and radius3 give $(x-2{)}^2+(y+1{)}^2=9$. The point $(5,-1)$ lies on the Euclidean circle; $(2,1)$ lies inside the closed disk because its distance from the center is2. Such an equation distinguishes a distance constraint from a region-filling inequality.

Every Euclidean circle is a circle in the topological sense. The map $u\mapsto c+ru$ from the unit circle $S^1$ is a homeomorphism, with inverse $x\mapsto (x-c)/r$. Topology forgets the particular center and radius, whereas the Euclidean distance retains them. The assumption $r>0$ matters: radius zero gives a point, which is not a circle.

The historical terminology differs slightly. Euclid's definition describes the plane figure contained by a circumference; the modern set above corresponds to that boundary. His equal-distance condition and distinguished center appear in BookI, Definitions15–18.^[ref]

Every continuous injective map $g:S^1\to S^1$ is a homeomorphism. After normalizing $g(1)=1$ by a rotation, it is homotopic either to the identity or to complex conjugation, and has degree $1$ or $-1$.

To see this without assuming surjectivity, lift $t\mapsto g(e^{2\pi it})$ on $[0,1]$ through the covering map $u\mapsto e^{2\pi iu}$ to a path $G$ with $G(0)=0$. Injectivity makes $G$ injective on $[0,1)$, hence strictly monotone. Its endpoint $d=G(1)$ is an integer. Strict monotonicity excludes $d=0$; if $|d|>1$, the intermediate value theorem gives an interior integer value, contradicting injectivity of $g$. Thus $d=\pm 1$, and $\exp (2\pi i((1-s)G(t)+sdt))$ descends to a homotopy from $g$ to $z\mapsto z^d$. If $g$ omitted a point, a continuous branch of the complex argument on the punctured circle would make it null-homotopic, contradicting this nonzero degree. Hence it is surjective. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

For real $A,C$ and complex $B$, if $A\ne 0$, completing the square gives $|z+B/A{|}^2=(|B{|}^2-AC)/A^2$. A positive right side gives a circle; zero gives a single point; a negative value gives the empty set. If $A=0$ and $B\ne 0$, the equation is the straight line $2\mathrm{Re}(\overline{B}z)+C=0$. If $A=B=0$, it describes the whole complex plane when $C=0$, and the empty set otherwise. Thus unrestricted coefficients require both the single-point and whole-plane degeneracies.

For the minor circular arc cut off by a chord of length $0\le L\le 2R$ on a circle of radius $R>0$, the sagitta is the maximum perpendicular distance from the chord to that arc:The perpendicular bisector of the chord passes through its midpoint and the center. The Pythagorean theorem gives the center-to-chord distance $\sqrt{R^2-(L/2{)}^2}$; subtracting it from $R$ proves the formula. The major circular arc instead gives $R+\sqrt{R^2-L^2/4}$, so choosing the arc matters.

For $L/R\ll 1$, a Taylor expansion gives $s=L^2/(8R)+O(L^4/R^3)$. In a uniform magnetic field perpendicular to a charged track's motion, $R=p/(|q|B)$, so the sagitta measures the momentum $p$.

Given three mutually externally tangent circles with disjoint interiors, a Soddy circle is a circle tangent to all three. The inner solution lies in the curvilinear triangular gap. In signed curvatures $b_i=\pm 1/r_i$, the two possible fourth curvatures satisfyA containing circle uses negative curvature. For three unit circles, the inner curvature is $3+2\sqrt{3}$, while the outer curvature $3-2\sqrt{3}<0$ describes a containing circle. The formula is the quadratic solution of the Descartes' circle theorem; zero curvature is interpreted as a tangent line.

Given three mutually externally tangent circles, the inner Soddy circle lies in their curvilinear triangular gap and is tangent to all three. The Descartes' circle theorem determines its radius from the three given radii.

The power of a point $P$ relative to a circle with center $O$ and radius $r$ is $O{P}^2-r^2$. If a line through $P$ meets the circle at $A$ and $B$, the product of the directed distances $PA$ and $PB$ equals this quantity.

Two distinct circles are tangent when they meet at exactly one point. For external tangency the distance between centers is the sum of the radii; for internal tangency it is their absolute difference. These relations turn many circle packing problems into exact distance equations.

Four positive-radius circles with pairwise disjoint interiors, successive external tangencies, and four distinct contact points have those contact points on a circle. Apply the reciprocal Möbius transformation $w=1/(z-T)$ at one contact $T$. The two circles through $T$ become disjoint parallel lines; the other two disks lie between them and avoid zero. Collinearity for tangent circles between parallel lines places the three finite transformed contacts on a straight line. That straight line misses zero: otherwise its segment across the strip would pass through zero, while that segment is contained in the two transformed closed disks. Its inverse is therefore a genuine circle through $T$, containing all four original contacts. Allowing internal tangency without the pictured disjoint-interior hypotheses admits collinear degeneracies instead.

Let two positive-radius circles have external tangency at $Q$ and lie in the closed strip between two parallel lines, touching opposite boundaries at $P,R$. Then $P,Q,R$ are collinear. To prove this, choose coordinates with the boundaries $x=0,d$, put $P=(0,0)$ and $R=(d,h)$, and write the centres as $O_1=(r_1,0)$, $O_2=(d-r_2,h)$. The contact point isThe closed line segment $PR$ is contained in the union of the two closed disks: its portions from $P$ to $Q$ and from $Q$ to $R$ are chords. The strip and external tangency hypotheses are essential to this particular formulation.

Two distinct positive-radius circles have internal tangency when their boundaries meet at exactly one point and one disk contains the other. Their centre separation equals the absolute difference of their radii.

Two distinct positive-radius circles have external tangency when their closed disks meet at exactly one point. The Euclidean distance between their centres equals the sum of their radii. The contact point divides the segment of centres internally in the ratio $r_1:r_2$.

At a point $T$ on a circle with centre $O$, the tangent line is perpendicular to the radius $OT$. For example, differentiating $|\mathbf{r}(s)-O{|}^2=R^2$ along a parametric curve on the circle gives $(\mathbf{r}-O)\cdot {\mathbf{r}}^{\prime }=0$, showing that the radial vector and tangent direction are orthogonal vectors. Conversely, the perpendicular straight line through $T$ has only that point in common with the circle: for a nonzero displacement $v$ along it, $|T+v-O{|}^2=R^2+|v{|}^2>R^2$.

A circular segment is the part of a disk bounded by a chord and one of the two circular arcs joining its endpoints. For radius $R$ and the chosen arc's central angle $0\le \varphi \le 2\pi$ in radians, its area is $R^2(\varphi -\sin \varphi )/2$. Subtract the signed central triangle of area $R^2\sin \varphi /2$ from the circular sector of area $R^2\varphi /2$. Angles below $\pi$ give the smaller segment and angles above $\pi$ the larger one.

Take a circle of radius $R$, a vertex $A$ on its circumference, and two chords from $A$ symmetric about the diameter through $A$. If their angle is $0\le \theta \le \pi$, choose the arc between their other endpoints that does not contain $A$. The area bounded by the two chords and that arc is $R^2(\theta +\sin \theta )$. To calculate it, set $A=(0,R)$ and the other endpoints to $(\pm R\sin \theta ,-R\cos \theta )$. Their triangle has area $R^2\sin \theta (1+\cos \theta )$, and the adjoining circular segment, of central angle $2\theta$, has area $R^2(\theta -\sin \theta \cos \theta )$. The sum gives the formula, including the limiting zero-area and full-disk cases.

A circular arc is a connected portion of a circle. For radius $R>0$ and subtended central angle $0<\theta <2\pi$ measured in radians, its arc length is $R\theta$. Two distinct endpoints determine two circular arcs: when they are not opposite, the minor arc has $\theta <\pi$ and the major arc has $\theta >\pi$. Opposite endpoints determine two semicircles. The straight line segment joining the endpoints is a chord, rather than the curved circular arc itself.

A circular sector is the portion of a disc between two radii and the connecting circle arc. If its radius is $R$ and its central angle is $\theta$ radians, its area isIndeed, polar coordinates give $A={\int }_0^{\theta }{\int }_0^{R}rdrd\varphi$. The full disc corresponds to $\theta =2\pi$, a semicircle to $\theta =\pi$, and a quarter-disc to $\theta =\pi /2$. In a solid with horizontal sector cross-sections, radius $R(z)$ and angle $\theta (z)$, its volume is $\frac{1}{2}\int \theta (z)R(z{)}^{2}dz$. Holding the angles and heights fixed while scaling every radius by a factor $s$ scales that volume by $s^2$.

Half of a circle, with endpoints at opposite ends of a diameter. For radius $R>0$, the upper semicircle in the complex plane is parametrized by $z=R{e}^{it}$, $0\le t\le \pi$, going counterclockwise from $R$ to $-R$.

A chord joins two points of a circle by a straight segment. If the circle has radius $r$ and the chord midpoint lies a distance $d$ from its center, the perpendicular bisector and Pythagorean theorem give the displayed length. The length exceeds that of an inscribed equilateral triangle precisely when $d<r/2$.

For a circle of radius $r$, a chord with minor central angle $0\le \theta \le \pi$ has lengthBisecting the isosceles triangle formed by its two radii gives a right triangle with hypotenuse $r$ and opposite side $L/2$, proving the formula.

If the perpendicular distance from the center to the chord is $d$, the same right triangle gives $L=2\sqrt{r^2-d^2}$. For $r=5,d=3$, the length is eight. The maximum chord is the diameter $2r$, obtained at $d=0$ or $\theta =\pi$.

The corresponding minor arc has length $r\theta$ when the angle is measured in radians. For small angles the chord is approximately that arc length, since $\sin (\theta /2)\sim \theta /2$, but for a nonzero angle the straight chord is strictly shorter than the arc.

For real $\alpha \ne 0,\gamma$ and complex $\beta \ne 0$, completing the square in $\alpha |z{|}^2-\beta \bar{z}-\bar{\beta }z+\gamma =0$ gives center $\beta /\alpha$ and radius $\sqrt{|\beta {|}^2-\alpha \gamma }/|\alpha |$, when the radicand is nonnegative. The displayed map is reflection in the line through $0$ and $\beta$: it preserves the complex modulus, fixes the center, and squares to the identity. Hence it maps the circle onto itself, including a radius-zero degeneration.

For two nonempty planar circles with centers $c_i$ and radii $r_i\ge 0$, the greatest Euclidean distance between their points is the displayed expression. The triangle inequality gives the upper bound. When the centers differ, choose the two boundary points outward along the line of centers to attain it; when they coincide, choose opposite radial directions. For any third nonempty circle and a point $u$ on it, $|z-w|\le |z-u|+|u-w|$ proves $m(C_1,C_2)\le m(C_1,C_3)+m(C_2,C_3)$. This is not a metric, since $m(C,C)=2r$ for a positive-radius circle.

A cyclic chain of distinct circles between two disjoint circle boundaries, each tangent to both boundaries and its two neighbours. In concentric coordinates with inner radius $\rho$ and outer radius $R$, the chain circles have radius $r=(R-\rho )/2$ and centre distance $d=(R+\rho )/2$. A simple disjoint chain with $n\ge 3$ closes when $r/d=\sin (\pi /n)$.

If a simple Steiner chain closes after $n\ge 3$ circles, rotating the starting circle between concentric boundaries gives another closed $n$-circle chain. Applying a Möbius transformation proves the corresponding statement for arbitrary boundary circles. Construction must stay in the same intervening family and proceed without reversing the chosen direction. Merely choosing arbitrary circles tangent to the preceding circle permits backtracking and does not assert closure.

For a Steiner chain between concentric circles, tangencies of neighbouring circles are midpoints of their centres and lie on the circle of radius $\sqrt{R\rho }$. Under a Möbius transformation, their locus is a generalized circle, which may be a line. An ordinary-circle conclusion requires the pole of the inverse transformation to avoid the original locus.

An equation $|az+b\overline{z}|=ux+v$, with $z=x+iy$ and real coefficients, becomes $(a+b{)}^2{x}^2+(a-b{)}^2{y}^2=(ux+v{)}^2$ after squaring. If $(a+b{)}^2-u^2=(a-b{)}^2>0$, completing the square gives a circle. The original locus is only the part where $ux+v\ge 0$. Checking this sign on the candidate circle prevents extraneous points introduced by squaring. This method combines the complex modulus with elementary conic geometry.

The radius of a circle or sphere is the Euclidean distance from its centre to its boundary. The diameter is twice the radius.

Pi is the ratio of a Euclidean circle's circumference to its diameter.

The Chudnovsky algorithm computes Pi to high precision by evaluating the Chudnovsky series with controlled truncation and rounding errors. A direct implementation updates the ratio of successive terms rather than recomputing every factorial. Large computations use binary splitting to combine rational partial sums through a balanced recursion tree, reducing expensive high-precision divisions. Roughly one term is needed per fourteen requested decimal digits, with additional guard precision and a bound on the remaining tail. Fast convergence of the series does not alone certify printed digits: finite-precision arithmetic errors must also be bounded.

Binary splitting evaluates a finite rational sum or a hypergeometric series by recursively dividing its term range into two nearly equal parts and combining exact integer data. The resulting balanced recursion tree postpones high-precision divisions and can exploit fast multiplication of similarly sized integers.

For ordinary rational partial sums represented by $(P,Q)$ and $(R,S)$, the exact merge isFor example, merging $1/2+1/3=5/6$ and $1/5+1/7=12/35$ yields $(5\cdot 35+12\cdot 6)/(6\cdot 35)=247/210$. Hypergeometric applications additionally carry products of successive-term ratio numerators and denominators to avoid rebuilding them at every node.

In the Chudnovsky algorithm, this organization makes large integer arithmetic efficient while evaluating the rational partial sum exactly before the final conversion and square-root factors. It does not bound the omitted infinite tail or floating-point rounding by itself: those need separate error estimates. The benefit is the balanced arithmetic organization, not a change in the value of the series.

The Chudnovsky series is the rapidly converging identityThe absolute ratio of successive summands tends to $1728/640320^3$, approximately $6.58\times 10^{-15}$, giving about fourteen new decimal digits per term asymptotically. Keeping only $k=0$ gives $\pi \approx 640320^{3/2}/(12\cdot 13591409)$. The factorial ratio, not a direct geometric approximation of a circle, controls the numerical convergence.^[ref]

The Wallis product isIts finite product is $W_n=2^{4n}(n!{)}^4/[(2n+1)((2n)!{)}^2]$. The Wallis integrals give $2W_n/\pi =I_{2n+1}/I_{2n}\to 1$, proving both convergence of the product and its value. The strict integral bounds also give $\pi n/(2n+1)<W_n<\pi /2$.

Ivan Niven's 1947 proof by contradiction constructs an integer strictly between zero and one.^[ref] Suppose $\pi =a/b$ for positive integers $a,b$. For a positive integer $n$, putEvery derivative of $f_n$ at zero is an integer: its coefficients have denominator $n!$, its lowest polynomial degree is $n$, and differentiation contributes the divisible factorial. The symmetry $f_n(\pi -x)=f_n(x)$ gives the same assertion at $\pi$.

Cancellation gives $F_n+F_n^{\prime \prime }=f_n$, henceThe integrand is positive in $(0,\pi )$, but $x(a-bx)=bx(\pi -x)\le b{\pi }^2/4$, soFor large $n$ this is impossible for an integer. Therefore Pi is an irrational number.

The circumference of a circle is its boundary arc length. For radius $r$, it equals $2\pi r$. This growing boundary length accounts for the radial factor in the Rayleigh distribution.