Inversion in a circle
Tonight's idea
Fix a circle with centre \(O\) and radius \(r\). The inverse of a point \(P \ne O\) is the point \(P'\) on the ray from \(O\) through \(P\) with \(OP \cdot OP' = r^2\).
A line not through \(O\) inverts to a circle through \(O\), and vice versa. Circles not through \(O\) invert to circles not through \(O\). Lines through \(O\) stay put; inverting twice gives \(P\) back.
Proof for the line. Let \(A\) be the foot of the perpendicular from \(O\) to \(\ell\), and \(P\) any other point of \(\ell\). Since \(OA \cdot OA' = OP \cdot OP' = r^2\), we have \(OA/OP = OP'/OA'\). Triangles \(OAP\) and \(OP'A'\) share the angle at \(O\) with the sides around it in proportion, so they are similar (\(A\) matching \(P'\)). Hence \(\angle OP'A' = \angle OAP = 90°\), and by Thales (circle theorems) \(P'\) lies on the circle with diameter \(OA'\). \(\square\)
Inversion is one to one, so curves touching at one point have images touching at one point: tangency survives.
Watch one: two touching circles and their common tangent
Circles \(\omega_1\), \(\omega_2\) touch at \(T\); a line touches them at \(A\) and \(B\). Claim: \(\angle ATB = 90°\).
Invert in any circle centred \(T\). Both circles pass through \(T\), so they become lines; they met only at \(T\), so their images meet nowhere: parallel lines. The line \(AB\) misses \(T\), so it becomes a circle through \(T\) touching both parallel lines, at \(A'\) and \(B'\), the ends of a diameter; so \(\angle A'TB' = 90°\) by Thales. And \(A'\) lies on ray \(TA\), \(B'\) on ray \(TB\), so \(\angle ATB = 90°\).
Drag lines and circles and watch their images move: mathdetours.com/demos/inversion.html.
Your turn
- The circle of inversion has radius 6. Where do points at distance 2, 3, 6, 9 and 36 from \(O\) go? Which points of the plane are fixed?
- Invert in the unit circle centred at the origin. (a) Explain why the formula is \((x, y) \mapsto \left(\dfrac{x}{x^2 + y^2},\ \dfrac{y}{x^2 + y^2}\right)\). (b) Find the image of the line \(x = 2\) (centre and radius), and check by inverting \((2, 2)\) directly.
- Use the formula from exercise 2 to show that the circle \(x^2 + y^2 = 2x\) (centre \((1, 0)\), radius 1) inverts to the line \(x = \tfrac12\). Why was a line to be expected?
- Circles \(\omega_1\) and \(\omega_2\) touch at \(T\), one inside the other. Invert in a circle centred at \(T\). (a) Why are the images two parallel lines? (b) A third circle touches both \(\omega_1\) and \(\omega_2\). What is its image, and what fixes its size? (c) Deduce that a chain of circles, each touching the next and all touching \(\omega_1\) and \(\omega_2\), inverts to a row of equal coins.
Stretch: Pappus's chain
In exercise 4, let \(m\) be the line through \(T\) and the centres, \(C_0\) the chain circle centred on \(m\), \(C_1\) the one touching \(C_0\), \(C_2\) the next, and so on. Show the centre of \(C_n\) sits exactly \(n\) diameters of \(C_n\) above \(m\). Hint: choose the inversion circle so that \(C_n\) is its own image.
Answers
- \(OP' = 36/OP\): distances 2, 3, 6, 9, 36 go to 18, 12, 6, 4, 1. The fixed points are exactly the points on the circle of inversion.
- (a) \(P'\) is on the same ray as \(P\) at distance \(1/OP\), so \(P' = P/OP^2\), and \(OP^2 = x^2 + y^2\). (b) The foot of the perpendicular is \(A = (2, 0)\), so \(A' = (\tfrac12, 0)\). The image is the circle with diameter \(OA'\): centre \((\tfrac14, 0)\), radius \(\tfrac14\). Check: \((2, 2)\) has \(x^2 + y^2 = 8\), so it goes to \((\tfrac14, \tfrac14)\), which is at distance \(\tfrac14\) from \((\tfrac14, 0)\).
- On the circle, \(x^2 + y^2 = 2x\), so the image has \(x' = \dfrac{x}{x^2 + y^2} = \dfrac{x}{2x} = \tfrac12\). The circle passes through \(O\), and circles through \(O\) become lines; the computation only says which line.
- (a) Both pass through \(T\), so both become lines; their only common point was \(T\), which has no image, so the lines have no common point: parallel. (b) It misses \(T\), so it becomes a circle, and it still touches both parallel lines, so its diameter is the width of the strip between them. (c) Every chain circle has the same image diameter, and touching circles have touching images: a row of equal coins across the strip.
Stretch. Draw a line through \(T\) cutting \(C_n\) at \(X\) and \(Y\); the product \(TX \cdot TY\) is the same for every such line (power of a point). Invert in the circle centred \(T\) with \(r^2 = TX \cdot TY\). Then every point \(X\) of \(C_n\) goes to the other point \(Y\) of \(C_n\) on the same line, so \(C_n\) is its own image, and its coin has its own diameter \(d_n\). The line \(m\) passes through \(T\), so it inverts to itself; \(C_0\) is symmetric about \(m\), so its coin is centred on \(m\). The coins stack along the strip, which is perpendicular to \(m\), so the \(k\)th coin is centred \(k\, d_n\) above \(m\). The \(n\)th coin is \(C_n\) itself. \(\square\) (Pappus of Alexandria proved this in the fourth century, without inversion.)