A Novel Method for Drawing a Circle Tangent to Three Circles Lying on a Plane by Straightedge,Compass,and Inversion Circles

TL;DR

This paper introduces a new geometric method using inversion and straightedge-compass constructions to draw a circle tangent to three given circles on a plane, avoiding Laguerre transformations.

Contribution

The paper presents a novel geometric approach employing inversion and tangent circles to construct a circle tangent to three given circles without advanced transformations.

Findings

Method successfully constructs tangent circle using inversion techniques.

Algorithm enables drawing with straightedge and compass only.

Survey of conformal mapping and inversion theory included.

Abstract

In this paper, we present a novel method to draw a circle tangent to three given circles lying on a plane. Using the analytic geometry and inversion (reflection) theorems, the center and radius of the inversion circle are obtained. Inside any one of the three given circles, a circle of the similar radius and concentric with its own corresponding original circle is drawn.The tangent circle to these three similar circles is obtained. Then the inverted circles of the three similar circles and the tangent circle regarding an obtainable point and a computable power of inversion (reflection) constant are obtained. These circles (three inverted circles and an inverted tangent circle)will be tangent together.Just,we obtain another reflection point and power of inversion so that those three reflected circles (inversions of three similar circles) can be reflections of three original circles, respectively. In such a case,the reflected circle tangent to three reflected circles regarding same new inversion system will be tangent to the three original ones. This circle is our desirable circle. A drawing algorithm is also given for drawing desirable circle by straightedge and compass. A survey of conformal mapping theory and inversion in higher dimensions is also accomplished. Although, Laguerre transformation might be used for solution of this problem, but we do not make use of this method. Our novelty is just for drawing a circle tangent to three given circles applying a tangent circle to three identical circles concentric with three given ones and then inverting them as original ones by compass and straightedge not any thing else.

Equations

$$x_{i}\mapsto\frac{r^{2}x_{i}}{\sum_{j=1}^{n}x_{j}^{2}}\$$

$$x\mapsto\frac{r^{2}x}{|x|^{2}}=y\mapsto\frac{k^{2}y}{|y|^{2}}=(\frac{k}{r})^{2}x\$$

$$\frac{\bar{z}}{|z|}.\frac{z}{|z|}=R^{2}=1\$$

$$OA.O\acute{A}=OB.O\acute{B}=OC.O\acute{C}=OD.O\acute{D}=OE.O\acute{E}=Of.O\acute{f}=k^2$$

$$OI.O\acute{A}=OH.O\acute{B}=OL.O\acute{C}=OM.O\acute{D}=OJ.O\acute{E}=OK.O\acute{f}={\acute{k}}^2$$

$$OI.O\acute{A}.OH.O\acute{B}=\acute{k}^{4}\$$

$$OI.OH.\frac{k^{2}}{OA}.\frac{k^{2}}{OB}=\acute{k}^{4}\ \Rightarrow\frac{OH.OI}{OA.OB}=\frac{OL.OM}{OC.OD}=\frac{OJ.OK}{OE.Of}=\frac{\acute{k}^{4}}{{k}^{4}}=M^{2}\$$

$$\frac{OH.OI}{OA.OB}=(\frac{AT_{2_{2}}}{AT_{1_{2}}})^{2},~{}~{}\frac{OL.OM}{OC.OD}=(\frac{AT_{2_{1}}}{AT_{1_{1}}})^{2}~{}~{}\frac{OJ.OK}{OE.Of}=(\frac{AT_{2_{3}}}{AT_{1_{3}}})^{2}\$$

$$(\frac{AT_{2_{1}}}{AT_{1_{1}}})^{2}=(\frac{AT_{2_{2}}}{AT_{1_{2}}})^{2}=(\frac{AT_{2_{3}}}{AT_{1_{3}}})^{2}=M^{2}\$$

$$\frac{OL.OM}{OC.OD}=(\frac{AT_{2_{1}}}{AT_{1_{1}}})^{2}=M^{2}=\frac{AO_{1}^{2}-R_{1}^{2}}{AO_{1}^{2}-R^{2}}\Rightarrow AO_{1}^{2}-R_{1}^{2}=M^{2}.AO_{1}^{2}-M^{2}.R^{2}\Rightarrow AO_{1}=\sqrt{\frac{R_{1}^{2}-M^{2}.R^{2}}{1-M^{2}}}\$$

$$AO_{1}^{2}-2R_{1}^{2}=-R^{2},~{}~{}AO_{2}^{2}-2R_{2}^{2}=-R^{2},~{}~{}AO_{3}^{2}-2R_{3}^{2}=-R^{2}\$$

$$C_{1}(x,y):~{}~{}x^{2}+y^{2}+a_{1}x+b_{1}y+c_{1}=0,~{}~{}C_{2}(x,y):~{}~{}x^{2}+y^{2}+a_{2}x+b_{2}y+c_{2}=0,~{}~{}C_{3}(x,y):~{}~{}x^{2}+y^{2}+a_{3}x+b_{3}y+c_{3}=0\$$

$$AO_{1}^{2}=(x_{0}+\frac{a_{1}}{2})^{2}+(y_{0}+\frac{b_{1}}{2})^{2},~{}~{}R_{1}^{2}=\frac{1}{4}(a_{1}^{2}+b_{1}^{2})-c_{1}\Rightarrow AO_{1}^{2}-2R_{1}^{2}=x_{0}^{2}+y_{0}^{2}+a_{1}x_{0}+b_{1}y_{0}+c_{1}+(c_{1}-\frac{1}{4}(a_{1}^{2}+b_{1}^{2}))\$$

$$AO_{2}^{2}-2R_{2}^{2}=x_{0}^{2}+y_{0}^{2}+a_{2}x_{0}+b_{2}y_{0}+c_{2}+(c_{2}-\frac{1}{4}(a_{2}^{2}+b_{2}^{2}))\$$

$$AO_{3}^{2}-2R_{3}^{2}=x_{0}^{2}+y_{0}^{2}+a_{3}x_{0}+b_{3}y_{0}+c_{3}+(c_{3}-\frac{1}{4}(a_{3}^{2}+b_{3}^{2}))\$$

$$K_{1}(x,y)=AO_{1}^{2}-2R_{1}^{2}+R^{2}=C_{1}(x,y)+(c_{1}-\frac{1}{4}(a_{1}^{2}+b_{1}^{2}))+R^{2}=0\$$

$$K_{2}(x,y)=AO_{2}^{2}-2R_{2}^{2}+R^{2}=C_{2}(x,y)+(c_{2}-\frac{1}{4}(a_{2}^{2}+b_{2}^{2}))+R^{2}=0\$$

$$K_{3}(x,y)=AO_{3}^{2}-2R_{3}^{2}+R^{2}=C_{3}(x,y)+(c_{3}-\frac{1}{4}(a_{3}^{2}+b_{3}^{2}))+R^{2}=0\$$

$$K_{3}-K_{2}:~{}~{}C_{3}(x,y)-C_{2}(x,y)+c_{3}-c_{2}+\frac{1}{4}(a_{2}^{2}-a_{3}^{2}+b_{2}^{2}-b_{3}^{2})=0\$$

$$K_{2}-K_{1}:~{}~{}C_{2}(x,y)-C_{1}(x,y)+c_{2}-c_{1}+\frac{1}{4}(a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2})=0\$$

$$K_{3}-K_{1}:~{}~{}C_{3}(x,y)-C_{1}(x,y)+c_{3}-c_{1}+\frac{1}{4}(a_{1}^{2}-a_{3}^{2}+b_{1}^{2}-b_{3}^{2})=0\$$

$$AO_{2}^{2}-AO_{1}^{2}=2\overline{O_{2}O_{1}}.\overline{M_{1}H_{2}},~{}~{}AO_{3}^{2}-AO_{2}^{2}=2\overline{O_{3}O_{2}}.\overline{M_{2}H_{3}},~{}~{}AO_{3}^{2}-AO_{1}^{2}=2\overline{O_{3}O_{1}}.\overline{M_{3}H_{1}}\$$

$$\overline{M_{1}H_{2}}=\frac{R_{2}^{2}-R_{1}^{2}}{\overline{O_{2}O_{1}}},~{}~{}\overline{M_{2}H_{3}}=\frac{R_{3}^{2}-R_{2}^{2}}{\overline{O_{3}O_{2}}},~{}~{}\overline{M_{3}H_{1}}=\frac{R_{3}^{2}-R_{1}^{2}}{\overline{O_{3}O_{1}}}\$$

$$(K_{3}-K_{2})+(K_{2}-K_{1})=(K_{3}-K_{1})\$$

Full text

A Novel Method for Drawing a Circle Tangent to

Three Circles Lying on a Plane by Straightedge,Compass,and Inversion Circles

Ahmad Sabihi

Professor and researcher at some of Iranian Universities Permanent Address: First floor, Bldg.name: Anita, Golbarg blind alley, Ershad alley (No.33), Avecina St., Isfahan, Iran, Tel:+98(31)34486478, Cell Phone:+989131115784, E-mail: sabihi2000@yahoo.com

**Abstract  In this paper, we present a novel method to draw a circle tangent to three given circles lying on a plane. Using the analytic geometry and inversion (reflection) theorems, the center and radius of the inversion circle are obtained. Inside any one of the three given circles, a circle of the similar radius and concentric with its own corresponding original circle is drawn.The tangent circle to these three similar circles is obtained. Then the inverted circles of the three similar circles and the tangent circle regarding an obtainable point and a computable power of inversion (reflection) constant are obtained. These circles (three inverted circles and an inverted tangent circle)will be tangent together.Just,we obtain another reflection point and power of inversion so that those three reflected circles (inversions of three similar circles) can be reflections of three original circles, respectively. In such a case,the reflected circle tangent to three reflected circles regarding same new inversion system will be tangent to the three original ones. This circle is our desirable circle. A drawing algorithm is also given for drawing desirable circle by straightedge and compass. A survey of conformal mapping theory and inversion in higher dimensions is also accomplished. Although, Laguerre transformation might be used for solution of this problem, but we do not make use of this method. Our novelty is just for drawing a circle tangent to three given circles applying a tangent circle to three identical circles concentric with three given ones and then inverting them as original ones by compass and straightedge not any thing else.

Keywords: Inversion Circles;Conformal Mapping;Transformation Geometry;Three-Circle Tangency Problem;Inversion in Higher Dimension; Laguerre transformation**

1 Introduction

The most important and interest problem,which was posed as a challenge by Apollonius [1] in the third century BC is three-circle tangency problem.This problem is stated as ”given three circles lying on a certain plane in general,construct all circles tangent to them ”. Apollonius generated the construction for nine cases,briefly, given three objects, each of them being a circle, a line, or a point, construct a circle tangent to all three given objects. The simplest case is the elementary constructs of a circle through three given points. The most challenging case is that one can draw a circle tangent to three given circles.That case is of the most emphasis here. Euclid and Archmedes had presented the construction for three points and three lines,but Apollonius gave the construction methods in his own books I and II on three-circle problem as well. Altshiller-Court [2]in 1925 gave a construction of same problem by reducing it to the construction of PCC (Point Circle-Circle). François Viète, Isaac Newton, and Joseph-Diaz Gergonne [3] individually solved the problem by different approaches.Viète made use of the geometric fundamentals, Newton the conic approach, and Gergonne developed a new understanding of inversion geometry properties.Viète’s work is fine for any three given circles,but not in an environment, where the given conditions can change after completion of the construction. Newton’s solution is also a bit problematic when it comes to dynamic geometry software. A change in the size or location of one of the given circles is likely to capsize the construction. His construction method is not unique so that decision must be made along the construction way. The method is based on the intersection of hyperbolas. An intersection of two hyperbolas is the case not three. Two hyperbolas can intersect at four points, and only two of which solve problem [3]. Gergonne in 1816 [3] published an inversion-based solution to the tangency problem. The Gergonne’s solution can be applied to many of the nine other cases.His method is the most exhaustive solution,which is specially flexible regarding positions of the given circles.

Inversion was invented by J.Steiner around 1830 and transformation by inversion in circle was invented by L.I.Magnus in 1831 [5]. For further information, refer to [6-9]. The author,surveys on inversive geometry,transformation theory, and conformal mapping in the Section 2. A novel geometrical method for drawing a circle tangent to the three given circles is given in the Section 3. A new algorithm for drawing same by Compass,and Straightedge without using any other axillary tools comes in the Section 4.

2 Inversive Geometry,Transformation Theory,and Conformal Mapping

Let $C_{1}$ be a circle passing through the point $A$ perpendicular to $C_{2}$. Let $\acute{A}$ be a second point of intersection of $OA$ with $C_{1}$,then $OC$ is tangent to $C_{1}$. This means that $OA.O\acute{A}=OC^{2}=R^{2}$ where $R$ is the radius of $C_{2}$. $C_{2}$ and $O$ are called the reflection circle and center respectively. Also,the location of point $\acute{A}$ does not depend on circle $C_{1}$,because $\acute{A}$ lies on the line $OA$ at distance equal to $\frac{R^{2}}{OA}$. The point $\acute{A}$ is known as the inverse,inversion,or reflection of $A$ with respect to $C_{2}$. Therefore,$\acute{A}$ is the reverse or reflect of $A$ and vice versa. Except the center of $C_{2}$,which is called as the center of inversion,all points as $A$ in the plane have inverse images as $\acute{A}$. The term inversion also applies to define transformation of the plane.This transformation maps the points inside $C_{2}$ to the points in its exterior and vice versa. The center of inversion is often [4-5] left over as a point with no inverse image,but sometimes is said to be mapped to the point at infinity. The origin or zero in the circle inversion mapping requires a special definition to adjoin a point at infinity as $\infty$. $R$ is called the radius of inversion,$R^{2}$ as the power of inversion. As stated in the introduction,inversive geometry refers to a study of transformation geometry by Euclidean transformations together with inversion in an n-sphere so:

[TABLE]

where $r$ denotes the radius of the inversion. In two dimensions,with $r=1$,this is indeed a circle inversion with respect to the unit circle so that $x\mapsto\frac{r^{2}x}{|x|^{2}}$. If two inversions in concentric circles are combined,then a similarity,homothetic transformation will be resulted by the ratio of the circle radii:

[TABLE]

The algebraic form of the inversion in a unit circle is $w=\frac{1}{\bar{z}}$ because

[TABLE]

and the reciprocal of $Z$ is $\frac{\bar{z}}{z}$. In the complex plane,reciprocation leads to the complex projective line and is often called the Riemann sphere. The subspaces and sub groups of this space and group of mappings,were applied to produce early models of hyperbolic geometry given by Arthur Cayley,Felix Klein, Henri Poincaré. Therefore,inverse geometry includes the ideas originated by Lobachevsky and Bolyai in their plane geometry. More generally, a map $f:U\longrightarrow V$ is called conformal or angle-preserving at $v_{0}$,if it preserves oriented angles between curves through $v_{0}$ with respect to their orientations. Conformal maps preserve both angles and shapes of infinitesimally small figures,but not necessarily their size. Some examples of conformal maps could be made from complex analysis. Let $A$ be an open subset of the complex plane,$\mathbb{C}$,then a function $g:A\longrightarrow\mathbb{C}$ is defined as a conformal map if and only if it is holomorphic and its derivation is everywhere non-zero on $A$.But if $g$ is anti-holomorphic (conjugate to a holomorphic function),it still preserves angles except it reverses their orientation. Conformal mapping application in electromagnetic potential,heat conduction,and flow of fluids are taken into considerations. The ideas are that the problems at hand with a certain geometry could be mapped into a problem with simpler geometry or a geometry which have already been solved. A such mappings are found by transformation [10]. For further study,refer to [11-12]. Laguerre transformation [13] might be used for solving our problem using a plane spanned by three points in a three dimensional space and transforming it into a horizontal plane, but we do not make use of such transformations. Also, methods of Cyclography might be applied to solve it. Our novelty is just for drawing a circle tangent to three given circles applying a tangent circle to three identical circles concentric with three given ones and then inverting them as original ones by compass and straightedge not any thing else.

3 The Novel Method

**3.1 General Descriptions

**Let $C_{1},C_{2}$,and $C_{3}$ be three given circles of optional radii $R_{1}<R_{2}<R_{3}$. Let’s draw three similar circles of radii $R$ concentric with each one of circles so that inequality $R<R_{1}<R_{2}<R_{3}$ holds. Adjoin the centers of circles together by straight-lines. Then draw perpendicular bisector of the created line-segments. The intersection point of perpendicular bisectors is indeed the center of tangent circle to the three similar circles of the radii $R$ since it is equidistance from the circles’ centers. Also, having equal radii,the distances of the intersection point from three circles are identical. This means that we able to draw a tangent circle to the three similar circles from the intersection point. Just, let the problem be solved. Using the reflection theorems and methods,the reflected objects (or reflected circles) of three similar circles of radii $R$ with respect to a fixed point (where will be obtained later by drawing) are obtained. In this paper,we carry out following steps 3.1.1 through 3.1.7 based on Fig.1

3.1.1Consider the three given circles $C_{1},C_{2}$,and $C_{3}$ of radii $R_{1},R_{2}$ and ,$R_{3}$, respectively.

3.1.2Draw the concentric circles of radii $R$ denoted by $\underline{C_{1}},\underline{C_{2}}$,and $\underline{C_{3}}$ (these symbols are not shown in Fig.1) inside the circles $C_{1},C_{2}$,and $C_{3}$, respectively.

3.1.3Draw the tangent circle $C_{4}$ to the three similar circles of the radii $R$, if their centers are not on a line

3.1.4Draw the reflected circles of the three similar circles of radii $R$ denoted by $\acute{C}_{1},\acute{C}_{2}$,and $\acute{C}_{3}$ with respect to the reflection center $O$

3.1.5Draw the reflected circle of the tangent circle $C_{4}$ i.e.$\acute{C}_{4}$. This reflected circle is also tangent to the reflected circles of the three similar circles of the radii $R$.

3.1.6Define a new center and a power of inversion so that the three reflected circles $\acute{C}_{1},\acute{C}_{2}$,and $\acute{C}_{3}$ (created by three similar circles of radii $R$) can be as the new reflected circles of $C_{1},C_{2}$,and $C_{3}$. The new reflection center is identical to the old one.

3.1.7Draw the reflection of the tangent circle $\acute{C}_{4}$ with respect to the new given center (or same old one) and new reflection constant (power of inversion value). This circle denotes ${C^{\prime\prime}}_{4}$ and is a desirable circle in this paper, where is tangent to the three given circles.

3.1.8If the centers of the three given circles (and three similar circles)$C_{1},C_{2}$,and $C_{3}$ are on a straight line,then we should draw a tangent line to the three similar circles instead of a tangent circle and obtain the reflected circle of this line and continue steps 3.1.5 to 3.1.7.

**3.2 Determination of the Center and Radius of Reflection Circle

**

Theorem1, Let we have the first reflection center $O$ referring Fig.1. We know the following relations hold:

[TABLE]

where $k^{2}$ denotes the power of inversion of the three similar circles. The points $\acute{A},\acute{B},\acute{C},\acute{D},\acute{E}$ and $\acute{f}$ denote reflected points for points $A$,$B$,$C$,$D$,$E$ and $f$ on the three reflection circles $\acute{C_{1}},\acute{C_{2}}$,and $\acute{C_{3}}$, respectively. Then, to make the reflected circles $\acute{C_{1}},\acute{C_{2}}$,and $\acute{C_{3}}$ as the reflections of the circles $C_{1},C_{2}$, and $C_{3}$, respectively, a new center of reflection can be defined so that lies on the previous one. Hence,the center $O$ should be the reflection center of both reflection systems so that:

[TABLE]

**Proof

**To have $O$ as the common center of both reflection systems:

[TABLE]

[TABLE]

where $M^{2}$ denotes the ratio of the second power of the inversions of both reflection systems. This is a constant value. It also denotes the ratio of the powers of the point $O$ with respect to concentric circles of radii $(R_{1},R),(R_{2},R)$,and $(R_{3},R)$. This means that ratio of the tangent line segments drawn from the point $O$ over both concentric circles $(R_{1},R),(R_{2},R)$,and $(R_{3},R)$ should be identical. The problem is that we wish to find geometrical location of all points whose powers are identical with respect to two given concentric circles. Referring Fig. 1 and Fig.2,we have the relations (5-3).

[TABLE]

Therefore,using (4-3) and (5-3)

[TABLE]

where $AT_{1_{1}}$ to $AT_{1_{3}}$ denote tangent lines drawn from the point $A$ or same $O$ (a typical reflection center) to the three similar circles of the radii $R$,respectively. Also, $AT_{2_{1}}$ to $AT_{2_{3}}$ denote tangent lines from same point to the circles $C_{1},C_{2}$,and $C_{3}$, respectively. For example, for circles $C_{1}$ and $\underline{C_{1}}$ (two concentric circles shown in Fig.2) of the center $O_{1}$ we have:

[TABLE]

The geometrical location of the point $A$ is a circle of the radius $AO_{1}$ and the center $O_{1}$. For the two other concentric circles $(C_{2},\underline{C_{2}})$,and $(C_{3},\underline{C_{3}})$, the method is similar. Due to $0<M^{2}<1$ we can choose an optional value e.g. $M^{2}=\frac{1}{2}$. We know $O_{2}$ and $O_{3}$ are the centers and $R_{2}$ and $R_{3}$, the radii of two other concentric circles, respectively, therefore regarding (7.3), we have

[TABLE]

Let the representing parametric equations of $C_{1},C_{2}$,and $C_{3}$ as follows:

[TABLE]

Let $A(x_{0},y_{0})$ be the mentioned typical reflection center. Using the equations (9-3),we obtain the circle centers $O_{1}(\frac{-a_{1}}{2},\frac{-b_{1}}{2})$,$O_{2}(\frac{-a_{2}}{2},\frac{-b_{2}}{2})$, and $O_{3}(\frac{-a_{3}}{2},\frac{-b_{3}}{2})$ then

[TABLE]

By the same method,we obtain

[TABLE]

[TABLE]

Theorem2, let $K_{1}(x,y)$, $K_{2}(x,y)$ and $K_{3}(x,y)$ denote the new circle’s equations eliminating zero indices as follows, then the center of inversion for three given circles is obtained by intersection of normal lines created by these circles.

[TABLE]

[TABLE]

[TABLE]

**Proof

**Consider

[TABLE]

[TABLE]

[TABLE]

The equations (16-3) to (18-3) are indeed the equations of the normal lines to the lines which adjoin the centers of both circles $(C_{3},C_{2})$, $(C_{2},C_{1})$, and $(C_{3},C_{1})$, respectively. Take the points $M_{1},M_{2}$ and $M_{3}$ placing at the bisector of the line segments constructed, then referring to Fig.3, we could make the relations (19-3).

[TABLE]

Therefore, the lines passing through the points $H_{1},H_{2}$,and $H_{3}$ denote our desirable geometrical locations of the points. These locations point out to the set of the points, whose difference of their power of distances from the centers of each two circles are constant values. These are the lines perpendicular to the line segments adjoining their two centers. The distance between the perpendicular line and the center of the line segment for each two circles is:

[TABLE]

The given lines presented by the equations (16-3) to (18-3) should intersect each other at a unique point because

[TABLE]

Therefore, from the intersection of the two lines $(K_{3}-K_{2})$ and $(K_{2}-K_{1})$,the line $(K_{3}-K_{1})$ is obtained.Thus,the point $(x,y)$ satisfies three line equations (16-3) through (18-3). This means that the center of inversion is found and the problem is solved.

4 The Novel Drawing Method by the Straightedge and Compass in form of a New Algorithm

The algorithm is given regarding Figs.1,3

4.1Draw three similar concentric circles $\underline{C_{1}},\underline{C_{2}}$,and $\underline{C_{3}}$ of radii $R$ inside the three given circles $C_{1},C_{2}$,and $C_{3}$, respectively so that $R<R_{1}<R_{2}<R_{3}$.

4.2Adjoin the centers of the circles $C_{1},C_{2}$,and $C_{3}$ to each other.

4.3Draw the perpendicular bisectors of the line segments adjoining each of two centers to meet at a point as ${O^{\prime\prime}}$(not shown in Fig.3).

4.4Adjoin the point ${O^{\prime\prime}}$ and center of one of the circles,say,$\underline{C_{3}}$ (Fig.3) to meet at the point $Z$.

4.5 Draw a circle of the center ${O^{\prime\prime}}$ and radius ${O^{\prime\prime}}Z$ to be tangent to the three similar circles $\underline{C_{1}},\underline{C_{2}}$,and $\underline{C_{3}}$.

4.6 Regarding Fig.3,the reflection center point $A$ could be found as follows:

Draw the radical axis of both circles $(C_{1},C_{2})$,$(C_{1},C_{3})$, and $(C_{2},C_{3})$.This can be done by drawing two optional circles to be intersected with circles,say,$C_{2}$,and $C_{3}$. These two radical axes intersect with one another at the point $\acute{O}$. A perpendicular line should be drawn from this point to the line segment $O_{2}O_{3}$ to meet it at a point. This point is called $\acute{H}$. Lie the compass’s needle on the $\acute{H}$ and open its other arm by $\acute{H}M_{2}$ based on what is shown in the Fig.3. Then, draw an arc toward smaller circle side,say,$C_{2}$, to be intersected with $O_{2}O_{3}$ at the point $H_{3}$. Draw a perpendicular line to $O_{2}O_{3}$ from the point $H_{3}$. This line is indeed representing the line $(K_{3}-K_{2})$. The method should be repeated for drawing the lines $(K_{2}-K_{1})$, and $(K_{3}-K_{1})$ by same method. These lines will intersect each other at the point $A$,which denotes the reflection center.

4.7 Obtain the reflections of three similar circles of equal radii $R$ (i.e. $\underline{C_{1}},\underline{C_{2}},\underline{C_{3}}$)and $C_{4}$ with respect to the reflection center $A$ and power of inversion $K^{2}$. These reflections denote $\acute{C_{1}},\acute{C_{2}},\acute{C_{3}}$,and $\acute{C_{4}}$ according to Fig.1. The circle $\acute{C_{4}}$ is tangent to the circles $\acute{C_{1}},\acute{C_{2}}$,and $\acute{C_{3}}$.

4.8According to given proof, three circles $\acute{C_{1}},\acute{C_{2}}$,and $\acute{C_{3}}$ can also be as reflections of the three circles $C_{1},C_{2}$,and $C_{3}$ with regard to same reflection center and by different power of inversion $\acute{K}^{2}$.

4.9Based on the explanations given in 4.8 and referring to the relation (4-3), we find out that $\frac{\acute{k}^{4}}{{k}^{4}}=M^{2}$. Therefore, the new power of inversion is obtained by the relation $\acute{k}^{2}=M.{k}^{2}$. Both ${k}^{2}$ and $M$ are known,thus $\acute{k}^{2}$ is also known, because, we can obtain size $M$ by the relation $M^{2}=\frac{1}{2}$ and by means of straightedge and compass tools. This means that the reflections of the three given circles $\acute{C_{1}},\acute{C_{2}}$,and $\acute{C_{3}}$ of the center $A$ and constant $\acute{k}^{2}$ can be the three main circles $C_{1},C_{2}$,and $C_{3}$, respectively.

4.10Referring to Fig.1,the reflection of the circle $\acute{C_{4}}$ is ${C^{\prime\prime}_{4}}$,which is a desirable solution. ${C^{\prime\prime}_{4}}$ is tangent to the three circles $C_{1},C_{2}$,and $C_{3}$ due to ${C^{\prime\prime}_{4}}$ is also reflection of $\acute{C_{4}}$ with regard to the point $A$ and the reflection constant $\acute{k}^{2}$.Therefore,the problem of three-circle tangency is completely drawn by compass and straightedge.

Only tools applying in this method to draw desirable circle,are both Compass and straightedge without any additional ones. Our method may be extended to three-dimension space by turning the plane of each one of circles about their related diameter. In such a case,we can make a sphere tangent to three given spheres. Also,this method would also be interested if it could be extended to higher dimension spheres. If three original circles are of concentric circles or only one of three circles is placed inside another one and the other one is outside of both,then we will not be able to draw a tangent circle to them. Further investigations about drawing limitations are left for readers.