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.

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 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 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 $\pi n/(2n+1)<W_n<\pi /2$.

Assuming $\pi =a/b$ and integrating a suitably normalized polynomial times $\sin x$ produces a positive integer that tends to zero as the polynomial degree grows. This contradiction proves that $\pi$ is an irrational number.