Properties of circles in mutual contact

PROPERTIES OF CIRCLES IN MUTUAL CONTACT. By Mr., Hyde, Cheshire.

Before entering upon the properties of four circles in mutual contact, it will be necessary to premise the following proposition; it is but a simple theorem in itself, and has place here only on account of its use in demonstrating the subsequent properties.

A. The continued product of the radii of any three circles mutually touching each other, divided by the radius of the circle passing through their points of contact, gives the area of the triangle formed by joining their centres.

Case 1. Let FEI, DFQ. DER, be three circles mutually touchlng each other exteriorly in the points D, E, F, and through their points ofcontact describe the circle DEF. Join the centres A, B, C, of the circles FEI, DFQ, DER; then the sides of the triangle thus formed will pass through the points D, E, F. Because AE = AF, BF = BD, CD = CE, the inscribed circle of the triangle ABC, will pass through the polnts D, E, F, and the circle DEF, passing through these points, is therefore the inscribed circle.

Denote the radii AF, BD, CE, of the circles touching each other by ρ$1$, ρ$2$, ρ$3$, respectively, and the radius of the circle passing through their points of contact by r; then by the Appendix to the Ladies' Diary, 1835, Davies's Horæ Geo. VII., Cor. 3, we have
 * $$\frac{\rho_1\rho_2\rho_3}{r}$$ = area of ΔABC, which gives the theorem in this case.

Case 2.&mdash;Let fei, dfq, der, be the position of the three circles in mutual contact, the circle fei touching the other two interiorally, and let def be the circle which passes through their points of contact. Join the centres a, b, c, of the circles fei, dfq, der, and produce the sides of the triangle thus formed; then in the same manner as above it may be shown that the circle def will touch the side bc of the triangle abc, and the sides ab, ac produced, in the points d, e, f, or def is the escribed circle to the side bc of the triangle abc. Hence (denoting the radii af, bd, ce, by ρ$1$, ρ$2$, ρ$3$, respectively, the radius of the circle def by r$1$, and the inscribed radius of the triangle abc by r), we shall have from Davies's Horæ Geo. II. and IV., ρ$1$ = ½{a + b + c}, and ρ$2$ρ$3$ = rr$1$
 * $$\begin{align}

\text{Therefore } \rho_1\rho_2\rho_3 = \rho_1rr_1 &= \textstyle\frac{1}{2}\{a + b + c\}rr_1 \\ &= \Delta abc\times r_1. \end{align} $$
 * Hence $$\frac{\rho_1\rho_2\rho_3}{r_1}$$ = Δabc,

which gives the theorem in this case; and these two cases include every position in which three circles can mutually touch one another.
 * $$\begin{align}

\textit{Cor.}\quad\text{In Case 1,} &\,\frac{1}{\rho_1\rho_2} + \frac{1}{\rho_2\rho_3} + \frac{1}{\rho_3\rho_1} = \frac{1}{r^2};\\ \text{and in Case 2,} &\,\frac{1}{\rho_2\rho_3} - \frac{1}{\rho_3\rho_1} - \frac{1}{\rho_1\rho_2} = \frac{1}{r_1^2}. \end{align} $$

The first of these equations has been established in Davies's Horæ Geo. VII., and the other follows from the relations
 * $$\rho_1\rho_2\rho_3 = r_1\times\Delta abc = r_1\times r_1 (\rho_1 - \rho_1 - \rho_3)$$ Horæ Geo. V. and VII.

I shall now proceed to discuss the properties of four circles in mutual contact; which from their remarkable agreement I shall designate, for the sake of reference,

The Concordant Circles.

I.

If any four circles be described to touch each other mutually, another set of four circles of mutual contact may be described whose points of contact shall coincide with those of the first four.

Let FdE, FeD, DfE, def be four circles mutually touching each other in the points D, E, F, d, e, f; through the points of contact of every three of these circles describe the circles DEF, Def, dEf, deF; then these four circles will also mutually touch each other in the points D, E, F, d, e, f. Join the centres A, B, C, P, of the four circles FdE, FeD, DfE, def, that were first described.

By Prop. A, the circles DEF, Def, dEf, deF, are respectively inscribed in the triangles ABC, PBC, PAC, PAB, and touch the sides of these triangles in the triads of points D, E, F, D, e, f,  d, E, f,  d, e, F, respectively. Hence AB being a common tangent to the two circles DEF, deF, at the point F, these two circles touch each other in this point; similarly the two circles deF, Def, touch each other in the point e, since BP is a common tangent to them at his point; also, by identical reasoning, all the four circles DEF, Def, dEf, deF, mutually touh each other in the points D, E, F, d, e, f.

Cor. The right lines AB, BC, CA, AP, BP, CP, joining the centres of the circles FdE, FeD, DfE, def, are separately tangents to two of the circles DEF, Def, dEf, deF, and conversely the lines ab, bc, ca, aO, bO, cO, joining the centres of these four last circles, are severally tangents to two of the former four; also one set of four circles respectively pass through the points of contact of every three of the other set.

II.

Denote the radii of the four cirles def, FdE, FeD, DfE, mutually touching each other by ρ, ρ$1$, ρ$2$, ρ$3$, respectively, and the radii of the circles DEF, Def, dEf, deF, touching each other in the same points as the other four by r, r$1$, r$2$, r$3$, respectively; then
 * $$\frac{1}{\rho} + \frac{1}{\rho_1} + \frac{1}{\rho_2} + \frac{1}{\rho_3} = -\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3},$$ the reciprocal of the radius of the circle touching three others interiorally having the negative sign.

For, we have from the corollary to Prop. A, the equations
 * $$\frac{1}{r^2} = \frac{1}{\rho_1\rho_2} + \frac{1}{\rho_2\rho_3} + \frac{1}{\rho_3\rho_1};\quad\frac{1}{r_1{}^2} = \frac{1}{\rho_2\rho_3} + \frac{1}{\rho_3\rho} + \frac{1}{\rho\rho_2};$$
 * $$\frac{1}{r_2{}^2} = {1\atop\rho_1\rho_3} + \frac{1}{\rho_3\rho} + \frac{1}{\rho\rho_1};\quad\frac{1}{r_3{}^2} = \frac{1}{\rho_1\rho_2} + \frac{1}{\rho_2\rho} + \frac{1}{\rho\rho_1};$$
 * $$\frac{1}{\rho^2} = \frac{1}{r_1r_2} + \frac{1}{r_2r_3} + \frac{1}{r_3r_1};\quad\frac{1}{\rho_1{}^2} = \frac{1}{r_2r_3} - \frac{1}{r_3r} - \frac{1}{rr_2};$$
 * $$\frac{1}{\rho_2{}^2} = \frac{1}{r_1r_3} - \frac{1}{r_3r} - \frac{1}{rr_1};\quad\frac{1}{\rho_3{}^2} = \frac{1}{r_1r_2} - \frac{1}{r_2r} - \frac{1}{rr_2}.$$

By adding together the first four equations, we get,
 * $$\frac{1}{r^2} + \frac{1}{r_1{}^2} + \frac{1}{r_2{}^2} + \frac{1}{r_3{}^2} = \frac{2}{\rho_1\rho_2} + \frac{2}{\rho_2\rho_3} + \frac{2}{\rho_3\rho_1} + \frac{2}{\rho\rho_1} + \frac{2}{\rho\rho_2} + \frac{2}{\rho\rho_3}.$$

By adding together the last four equations, we have,
 * $$\frac{2}{r_1r_2} + \frac{2}{r_2r_3} + \frac{2}{r_3r_1} - \frac{2}{rr_1} - \frac{2}{rr_2} - \frac{2}{rr_3} = \frac{1}{\rho^2} + \frac{1}{\rho_1{}^2} + \frac{1}{\rho_2{}^2} + \frac{1}{\rho_3{}^2}.$$

Adding this equation to the preceding one, we obtain,
 * $$\begin{align}

\frac{1}{r^2} + \frac{1}{r_1{}^2} + \frac{1}{r_2{}^2} + \frac{1}{r_3{}^2} + \frac{2}{r_1r_2} + \frac{2}{r_2r_3} + \frac{2}{r_3r_1} - \frac{2}{rr_1} - \frac{2}{rr_2} - \frac{2}{rr_3}\\ = \frac{1}{\rho^2} + \frac{1}{\rho_1{}^2} + \frac{1}{\rho_2{}^2} + \frac{1}{\rho_2{}^2} + \frac{2}{\rho_1\rho_2} + \frac{2}{\rho_2\rho_3} + \frac{2}{\rho_3\rho_1} + \frac{2}{\rho\rho_1} + \frac{2}{\rho\rho_2} + \frac{2}{\rho\rho_3} \end{align}$$
 * or $$\Big(-\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\Big)^2 = \Big(\frac{1}{\rho} +\frac{1}{\rho_1} + \frac{1}{\rho_2} + \frac{1}{\rho_3}\Big)^2$$

which gives the theorem.

This neat theorem may be expressed in the following manner:

The sum of the reciprocals of the radii of four circles mutually touching one another is equal to the sum of the reciprocals of the radii of the four other circles mutually touching each other in the same points. Observing that when the radii of those circles which respectively touch three others exteriorally are treated as positive quantities, the radii of those circles which respectively touch three others interiorally must be considered as negative quantities; and when this is taken into consideration we shall have this and the following theorems general for all positions of four circles in mutual contact.

III.

To determine the radii of four circles mutually touching each other, in terms of the radii of the four circles mutually touching each other in the same points as the other four.

From the first four in Prop. , we get


 * $$\begin{align}

-\frac{1}{r^2} + \frac{1}{r_1{}^2} + \frac{1}{r_2{}^2} + \frac{1}{r_3{}^2} &= \frac{2}{\rho}\Big\{\frac{1}{\rho_1} + \frac{1}{\rho_2} + \frac{1}{\rho_3}\Big\} \\ &= \frac{2}{\rho}\Big\{-\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} - \frac{1}{\rho}\Big\} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\text{From Prop. II} \\ &= \frac{2}{\rho}\Big\{-\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\Big\} - \frac{2}{\rho^2} &\\ &= \frac{2}{\rho}\Big\{-\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\Big\} - &\\ &\quad\quad\;\;\Big\{\frac{2}{r_1r_2} + \frac{2}{r_2r_3} + \frac{2}{r_3r_1}\Big\}\;\text{From Prop. II} \end{align}$$
 * $$\therefore\frac{2}{\rho}=\frac

{\displaystyle-\frac{1}{r^2} + \frac{1}{r_1{}^2} + \frac{1}{r_2{}^2} + \frac{1}{r_3{}^2} + \frac{2}{r_1r_2} + \frac{2}{r_2r_3} + \frac{2}{r_3r_1}} {\displaystyle-\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}}$$
 * or $$\frac{2}{\rho} = \frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}.$$

In a similar manner, we obtain
 * $$\begin{align}

\frac{2}{\rho_1} &= -\frac{1}{r} - \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \\ \frac{2}{\rho_2} &= -\frac{1}{r} + \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_3} \\ \frac{2}{\rho_3} &= -\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} - \frac{1}{r_3}. \end{align}$$

Obs. The first of these equations gives the following neat theorem:

If three circles be inscribed within another given one so as mutually to touch it and each other, the reciprocal of the radius of the circle passing through their points of contact, is equal to half the sum of the reciprocals of the radii of the four circles mutually touching each other.

Cor. 1.
 * $$\begin{align}

\frac{2}{r} = \frac{1}{\rho} - \frac{1}{\rho_1} - \frac{1}{\rho_2} - \frac{1}{\rho_3},\quad& \frac{2}{r_1} = \frac{1}{\rho} - \frac{1}{\rho_1} + \frac{1}{\rho_2} + \frac{1}{\rho_3},\\ \frac{2}{r_2} = \frac{1}{\rho} + \frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{1}{\rho_3},\quad& \frac{2}{r_3} = \frac{1}{\rho} + \frac{1}{\rho_1} + \frac{1}{\rho_2} - \frac{1}{\rho_3}. \end{align}$$ Cor. 2. $$\frac{1}{\rho} - \frac{1}{r} = \frac{1}{\rho_1} + \frac{1}{r} = \frac{1}{\rho_2} + \frac{1}{r_2} = \frac{1}{\rho_3} + \frac{1}{r_3}.$$

IV.

The sum of the squares of the reciprocals of the radii of four circles mutually touching one another, is equal to twice the sum of their rectangles.

For, from Prop. Cor. 1,
 * $$\Big(\frac{1}{\rho} - \frac{1}{\rho_1} - \frac{1}{\rho_2} - \frac{1}{\rho_3}\Big)^2 = \frac{4}{r^2} = \frac{4}{\rho_1\rho_2} + \frac{4}{\rho_2\rho_3} + \frac{4}{\rho_3\rho_1},$$ from.
 * or $$\frac{1}{\rho^2} + \frac{1}{\rho_1{}^2} + \frac{1}{\rho_2{}^2} + \frac{1}{\rho_3{}^2} + \frac{2}{\rho_1\rho_2} + \frac{2}{\rho_2\rho_3} + \frac{2}{\rho_3\rho_1} - \frac{2}{\rho\rho_1} - \frac{2}{\rho\rho_2}$$
 * $$- \frac{2}{\rho\rho_3} = \frac{4}{\rho_1\rho_2} + \frac{4}{\rho_2\rho_3} + \frac{4}{\rho_3\rho_1}$$
 * $$\therefore \frac{1}{\rho^2} + \frac{1}{\rho_1{}^2} + \frac{1}{\rho_2{}^2} + \frac{1}{\rho_3{}^2} = \frac{2}{\rho_1\rho_2} + \frac{2}{\rho_2\rho_3} + \frac{2}{\rho_3\rho_1} + \frac{2}{\rho\rho_1} + \frac{2}{\rho\rho_2} + \frac{2}{\rho\rho_3}.$$

When one of the circles touches the other three interiorally, its radius must be treated as a negative quantity in this theorem.

For from. $$\Big(\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\Big)^2 = \frac{4}{\rho^2} = \frac{4}{r_1r_2} + \frac{4}{r_2r_3} + \frac{4}{r_3r_1},$$
 * $$\therefore\frac{1}{r^2} + \frac{1}{r_1{}^2} + \frac{1}{r_2{}^2} + \frac{1}{r_3{}^2} = \frac{2}{r_1r_2} + \frac{2}{r_2r_3} + \frac{2}{r_3r_1} - \frac{2}{rr_1} - \frac{2}{rr_2} - \frac{2}{rr_3},$$

in which the radius (r), of the circle DEF, touching three others interiorally is negative.

V.

Given three circles mutually touching each other, to determine the radius of a circle touching all three in terms of the given circles.

This will be very readily done from Prop. ; for denoting the radii of the three given circles by $$R_\prime$$, $$R_{\prime\prime}$$, $$R_{\prime\prime\prime}$$, respectively, and the raidus of the required circle by R, we shall have
 * $$\begin{align}

\frac{1}{R^2} + \frac{1}{R_\prime{}^2} + \frac{1}{R_{\prime\prime}{}^2} + &\frac{1}{R_{\prime\prime\prime}{}^2} = \frac{2}{R_\prime R_{\prime\prime}} + \frac{2}{R_{\prime\prime} R_{\prime\prime\prime}} + \frac{2}{R_{\prime\prime\prime} R_\prime} + \frac{2}{RR_\prime} \\ &+ \frac{2}{RR_{\prime\prime}} + \frac{2}{RR_{\prime\prime\prime}}. \end{align}$$ Whence, $$\frac{1}{R} = \frac{1}{R_\prime} + \frac{1}{R_{\prime\prime}} + \frac{1}{R_{\prime\prime\prime}} \pm2\sqrt{\frac{1}{R_\prime R_{\prime\prime}} + \frac{1}{R_{\prime\prime} R_{\prime\prime\prime}} + \frac{1}{R_{\prime\prime\prime} R_\prime}}.$$

In which the double sign &plusmn; shows that there may be two circles described respectively to touch all the given three, the upper sign (+) determining the radius of the lesser circle, and the lower one (&minus;) the radius of the greater circle. And this equation will give the radius of the required circle of contact, when the three given circles touch each other in any position, by considering the radius of a circle touching the others interiorally to be a negative quantity. Thus, if the circle whose radius is $$R_\prime$$, touches the other circles interiorally,
 * $$\frac{1}{R} = \frac{1}{R_\prime} + \frac{1}{R_{\prime\prime}} + \frac{1}{R_{\prime\prime\prime}} \pm2\sqrt{\frac{1}{R_{\prime\prime} R_{\prime\prime\prime}} - \frac{1}{R_\prime R_{\prime\prime}} - \frac{1}{R_\prime} R_{\prime\prime\prime}}.$$

When the three given circles touch each other exteriorally, the larger of the two circles of contact will touch them interiorally, and its radius will therefore be a negative quantity, and the above equation for its determination then becomes
 * $$-\frac{1}{R} = \frac{1}{R_\prime} + \frac{1}{R_{\prime\prime}} + \frac{1}{R_{\prime\prime\prime}} - 2\sqrt{\frac{1}{R_\prime R_{\prime\prime}} + \frac{1}{R_{\prime\prime} R_{\prime\prime\prime}} + \frac{1}{R_{\prime\prime\prime} R_\prime}},$$
 * or, $$\frac{1}{R} = -\frac{1}{R_\prime} - \frac{1}{R_{\prime\prime}} - \frac{1}{R_{\prime\prime\prime}} + 2\sqrt{\frac{1}{R_\prime R_{\prime\prime}} + \frac{1}{R_{\prime\prime} R_{\prime\prime\prime}} + \frac{1}{R_{\prime\prime\prime} R_\prime}}.$$

(To be continued.)

$$\because$$ Mr. Beecroft's paper on Infinite Series will be inserted in our next.