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  "body": "\n## Proof\n\nWe write the map as a composition of simple maps and check each simple map.\nFor $z \\in \\C$, $\\bar z$ is the [[complex conjugate|0BJV24C4SY]] and $\\abs{z}$ is the [[modulus|0KW2X0TFBX]], so $z \\bar z = \\abs{z}^2$.\nGeneralized circles live in the [[Riemann sphere|0CDMXGPSES]] $\\widehat{\\C} = \\C \\cup \\set{\\infty}$.\n\n*Equations of generalized circles.*\n\n1. Call a triple $(A, B, D)$ admissible if $A, D \\in \\R$, $B \\in \\C$ and $\\abs{B}^2 \u003e AD$. Let $Z(A, B, D)$ be the set of $z \\in \\C$ with $A z \\bar z + B z + \\bar B \\bar z + D = 0$.\n2. Let $\\Gamma(A, B, D) = Z(A, B, D)$ if $A \\ne 0$, and $\\Gamma(A, B, D) = Z(A, B, D) \\cup \\set{\\infty}$ if $A = 0$. We show that the generalized circles are exactly the sets $\\Gamma(A, B, D)$ with $(A, B, D)$ admissible.\n3. Let $C$ be a [[circle|045DN06F3S]] with centre $c$ and radius $r \u003e 0$. Expanding $\\abs{z - c}^2 = r^2$ gives $z \\bar z - \\bar c z - c \\bar z + \\abs{c}^2 - r^2 = 0$. So $C = \\Gamma(1, -\\bar c, \\abs{c}^2 - r^2)$, and $\\abs{-\\bar c}^2 - (\\abs{c}^2 - r^2) = r^2 \u003e 0$.\n4. Let $L = \\set{p + t v : t \\in \\R}$ be a [[line|0BMYPJBJH4]], with $p \\in \\C$ and $v \\in \\C \\setminus \\set{0}$. A point $z$ lies on $L$ exactly when $(z - p)/v$ is real.\n5. Since $(z - p)/v = \\bar v (z - p)/\\abs{v}^2$, this holds exactly when $\\bar v (z - p) - v(\\bar z - \\bar p) = 0$. Multiply by $i$ and put $B = i \\bar v$, so $\\bar B = -iv$. The condition becomes $B z + \\bar B \\bar z + D = 0$ with $D = -(Bp + \\bar B \\bar p)$.\n6. Here $D = -2 \\operatorname{Re}(Bp)$ is real and $\\abs{B}^2 = \\abs{v}^2 \u003e 0$. So $L \\cup \\set{\\infty} = \\Gamma(0, B, D)$ with $(0, B, D)$ admissible.\n7. Conversely, let $(A, B, D)$ be admissible with $A \\ne 0$. Put $c = -\\bar B / A$. Dividing the equation by $A$ and using $\\abs{z - c}^2 = z \\bar z + (B/A) z + (\\bar B/A) \\bar z + \\abs{B}^2/A^2$, the equation becomes $\\abs{z - c}^2 = (\\abs{B}^2 - AD)/A^2$.\n8. The right side of step 7 is positive. So $\\Gamma(A, B, D)$ is the circle with centre $c$ and radius $\\sqrt{\\abs{B}^2 - AD}/\\abs{A}$.\n9. Let $(0, B, D)$ be admissible, so $B \\ne 0$. Put $v = i \\bar B$, so $v \\ne 0$ and $B = i \\bar v$. Put $p = -D \\bar B / (2 \\abs{B}^2)$, so $B p = -D/2$ and $D = -(Bp + \\bar B \\bar p)$.\n10. By steps 4 and 5, read backwards with these $v$ and $p$, the set $Z(0, B, D)$ is the line $\\set{p + tv : t \\in \\R}$. So $\\Gamma(0, B, D)$ is a line together with $\\infty$.\n11. Steps 3, 6, 8 and 10 prove the claim of step 2.\n\n*Affine maps.*\n\n12. Let $g(z) = \\alpha z + \\beta$ for $z \\in \\C$, with $\\alpha \\ne 0$, and $g(\\infty) = \\infty$. Then $g$ is a bijection of $\\widehat{\\C}$, with inverse $w \\mapsto (w - \\beta)/\\alpha$ on $\\C$.\n13. Let $C$ be the circle with centre $c$ and radius $r$. For $z \\in \\C$, $\\abs{g(z) - g(c)} = \\abs{\\alpha} \\abs{z - c}$. So $w \\in \\C$ satisfies $\\abs{w - g(c)} = \\abs{\\alpha} r$ exactly when $g^{-1}(w) \\in C$.\n14. By steps 12 and 13, $g(C)$ is the circle with centre $g(c)$ and radius $\\abs{\\alpha} r \u003e 0$.\n15. Let $L = \\set{p + tv : t \\in \\R}$ with $v \\ne 0$. Then $g(p + tv) = g(p) + t \\alpha v$ and $\\alpha v \\ne 0$. With $g(\\infty) = \\infty$, the image $g(L \\cup \\set{\\infty})$ is the line through $g(p)$ with direction $\\alpha v$, together with $\\infty$.\n16. So $g$ maps each generalized circle to a generalized circle.\n\n*The inversion.*\n\n17. Let $J(z) = 1/z$ for $z \\in \\C \\setminus \\set{0}$, $J(0) = \\infty$ and $J(\\infty) = 0$. Then $J$ is a bijection of $\\widehat{\\C}$ and $J \\circ J$ is the identity.\n18. Let $C = \\Gamma(A, B, D)$ with $(A, B, D)$ admissible, as in step 11. Then $(D, \\bar B, A)$ is admissible, since $\\abs{\\bar B}^2 = \\abs{B}^2 \u003e DA$. Let $C' = \\Gamma(D, \\bar B, A)$, a generalized circle by step 11.\n19. Let $z, w \\in \\C \\setminus \\set{0}$ with $w = 1/z$. Multiplying $A z \\bar z + B z + \\bar B \\bar z + D = 0$ by $w \\bar w \\ne 0$ gives $A + B \\bar w + \\bar B w + D w \\bar w = 0$. This is the equation of $Z(D, \\bar B, A)$.\n20. Since $w \\bar w \\ne 0$, the two equations of step 19 are equivalent. So $J$ maps $C \\setminus \\set{0, \\infty}$ onto $C' \\setminus \\set{0, \\infty}$.\n21. Putting $z = 0$ in the equation shows $0 \\in C$ exactly when $D = 0$. By step 2, $\\infty \\in C'$ exactly when $D = 0$. So $0 \\in C$ exactly when $J(0) = \\infty \\in C'$.\n22. By step 2, $\\infty \\in C$ exactly when $A = 0$. Putting $w = 0$ in the equation of $C'$ shows $0 \\in C'$ exactly when $A = 0$. So $\\infty \\in C$ exactly when $J(\\infty) = 0 \\in C'$.\n23. By steps 20, 21 and 22, $J(C) = C'$. So $J$ maps each generalized circle to a generalized circle.\n\n*Decomposition.*\n\n24. Let $f(z) = \\frac{az + b}{cz + d}$ be a [[Möbius transformation|0B346DDAM5]], with $ad - bc \\ne 0$.\n25. If $c = 0$, then $ad \\ne 0$, so $d \\ne 0$. Then $f(z) = (a/d) z + b/d$ on $\\C$ and $f(\\infty) = \\infty$, with $a/d \\ne 0$. So $f$ is a map $g$ as in step 12.\n26. Let $c \\ne 0$. Put $g_1(z) = cz + d$ and $g_2(w) = a/c + k w$ with $k = (bc - ad)/c \\ne 0$, both extended by $\\infty \\mapsto \\infty$.\n27. For $z \\in \\C$ with $cz + d \\ne 0$: $g_2(J(g_1(z))) = \\frac{a}{c} + \\frac{bc - ad}{c(cz + d)} = \\frac{a(cz + d) + bc - ad}{c(cz + d)} = \\frac{az + b}{cz + d} = f(z)$.\n28. At $z = -d/c$: $g_1(z) = 0$, $J(0) = \\infty$ and $g_2(\\infty) = \\infty = f(-d/c)$. At $z = \\infty$: $g_1(\\infty) = \\infty$, $J(\\infty) = 0$ and $g_2(0) = a/c = f(\\infty)$.\n29. By steps 27 and 28, $f = g_2 \\circ J \\circ g_1$ on all of $\\widehat{\\C}$.\n30. Let $C$ be a generalized circle. By steps 16 and 23, applied in turn, $g_1(C)$, $J(g_1(C))$ and $g_2(J(g_1(C))) = f(C)$ are generalized circles. With step 25 this covers both cases, so $f(C)$ is a generalized circle.\n\n## Idea\n\nEvery circle and every line is the zero set of one equation $A \\abs{z}^2 + Bz + \\bar B \\bar z + D = 0$. Substituting $z = 1/w$ and clearing denominators gives an equation of the same shape with $A$ and $D$ swapped, so the inversion keeps the family. Affine maps multiply all distances by $\\abs{\\alpha}$, so they keep circles and lines, and every Möbius map is built from these pieces.\n",
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