Now we show that the inversion of two points can be induced with the help of circles of equal radius.

Draw an arbitrary circle c.

Fig. 1

Point on the circumference c two points A and B, incident to this circle and draw two circles a and b with centers at these points so that their radii are the same.

Fig. 2

Connect the centers of the circles A and B withf the line AB.

Fig. 3

Find the point of intersection of the circles a and b to c and construct a circle intersecting ta and tb.

Fig. 4

Define the intersection point Q intersecting ta and tb.

Fig. 5

To establish the connection between the original data and the desired point of inversion construct the image of the point Q in the inversion circle c - point P.

Fig. 6

Considering the point C of the circle center point Q c and a diametrical construct them circle d.

Fig. 7

Define the intersection points A 'and B' with the circumference of the circle d c.

Fig. 8

Connect the points A 'and B' with straight line A'B '.

Line A'B 'cross the line l, connecting the center of the circle c - C point with the point Q at the point P.

Fig. 10

Construct the radical axis of the circles ra a and d, and the radical axis of the circles rb b, and d. Draw attention to the fact that the radical axis ta ra and intersect at a point that is incident to A'B '. A similar statement can be made with respect to the radical axis tb and rb.

Let us look at an arbitrary point of the plane S and draw through it and point B1 and B2 circle j. B1 and B2 points are centers of circles of the beam, each of which forms a circle with, as a representative of the same beam axis tb overall radical. Hence any of the circumferences of the beam except for circle c, can be used to set the same inversion points, it is obvious that is not necessarily the center circle should be chosen on a circle c.

Fig. 13

Thus, we can conclude that the same inversion points define two circles beam, which is common circle inversion element. Find the radical axis of the circles j and a. For this, as an example, we find the complex conjugate imaginary point of intersection of these circles p2 and p3 and draw through them the real line x line. As seen from the drawing, x-axis passes through the point Q.

Fig. 14

Now we construct a circle v, extending orthogonal to c and j through P. This circle will pass also through the point Q.

Fig. 15

Let us now consider such an option arrangement of the starting circles in which their centers are still on the circle c, but do not have the same radius values. Show the equivalence principle of the method of the initial data in the previous case.

Start a new project.

So let on the circle c are given two points A and B.

Fig. 16

Construct on them as centers, d1 and d2 of the circle, giving them unequal radii.

Fig. 17

Define two circumferential inversion d3 and d4, d1 and d2 transform into each other. To do this select circle d1 and d2, and press the key on the keyboard with the Latin symbol R (upper case). If the circles d1 and d2 do not overlap obviously, one of the circles is the inversion of the imaginary (in the figure is a circle d3).

Fig. 18

Construct anywhere on the plane an arbitrary straight line o1.

Fig. 19

Now we define the circle of inversion d5 and d6, d3 transform the circle and straight line o1 each other. As in the previous case, select d3 and o1 and press the key on the keyboard with the symbol R.

Fig. 20

Carry out the transformation of raw data on d5 circle in the transformation of inversion. To do this, select all objects by pressing the key combination Ctrl + A, then hold down the Shift key, exclude from the allocation of a circle and a straight line d5 o1, then re-add a circle d5 so that it is in the list of selected objects turned out to be the last. Now, pressing the key on the keyboard with the Latin symbol i, will carry out the transformation of inversion.

From the figure it is seen that the circumference of d1 and d2 transformed into p4 and p5 circle of the same radius, but the centers of these circles do not coincide with the images of the centers of circles starting p1 and p3. It is clear that the circumference of p4 and p5 are representatives of two circular beams which are symmetric with respect to the axis of the circles radical p4 and p5, therefore, essentially equivalent to the previous task has become.

Fig. 21

Find the point of intersection of p9 radical axes o2 and o3.

Fig. 22

Find the image p10 of point p9 relative p2 and draw circle, passing through p9 and p10 line o4.

Fig. 23

Transform p9, p10 and o4 in inversion d5. Get the point d8, d9 and circle d7.

Note that the point d8 and d9 are not in transforming respective inversion circle c. Point d9 corresponds to a point p13, which is the result of the intersection of intersecting o5 and o6. However, it should be noted that d9, p13, p14 and point p9 lie on a single straight line o7.

Fig. 25

Start a new project. Take an arbitrary circle d1, an arbitrary straight line o1 and assing two points p1 and p2 on it.

Fig. 26

Construct centers in p1 and p2 be two arbitrary circle d2 and d3.

Fig. 27

Define the intersection point p7 radical axes o2 and o3.

Fig. 28

Find the image p8 of point p7 in inversion d1.

Fig. 29

By specifying an arbitrary point p8 and taking as circles bundle center point p3 and p4, we construct d4 as a representative of the bundle.

Fig. 30

Similarly, we construct a circle d5, after spreading it through an arbitrary point p10 and the points p5 and p6.

Fig. 31

Draw the circumference d7 Tthrough the point p7, orthogonal to d3 and d4, and the circumference d6, d2 and orthogonal to d5. As seen from the drawing, the two circles intersect at a point and d8.

Fig. 32

Start a new project.

Let now it is given an arbitrary circle d1 and two arbitrary lines o1 and o2. The drawing lines clearly do not intersect d1.

Fig. 33

Find the intersection point p1 and p2 of line o1 the circle d1 and p3 and p4 intersection point of a straight line o2 with the circle d1.

Fig. 34

Pointing out anywhere an arbitrary point p5, draw the circle d2 passing through it and through the points p3 and p4 .

Fig. 35

Pointing out anywhere an arbitrary point p6, draw the circle d3 passing through it and through the points p1 and p2.

Fig. 36

Determine the intersection point p7 of lines o1 and o2 and find its image in the inversion on circle d1.

Fig. 37

Draw the circle d4 passing through the point p7 and perpendicular to d3 and o1.

Fig. 38

Draw the circle d5 passing through the point p7 and perpendicular to d2 and o2.

As seen from the drawing, p8 is located in the intersection of circles d4 and d5.

Fig. 39

Now consider the geometric design matches the inverse images of the points, if consistently consider three circles, as each of them acts as a circle inversion, and the other two point transformation induced.

Start a new project. Define three circles d1, d2 and d3 on a plane.

Fig. 40

Find the radical center of three circles - the point p1. To do this, select the three circle and press the key on the keyboard with the Latin symbol R (upper case).

Fig. 41

Transform radical center p1 in inversion relative to each of the circles d1, d2 and d3. Obtain, respectively points p2 , p3 and p4.

Fig. 42

Form a circle d4, d5 and d6, passing them through radical center and each pair of them received triple of points.

Fig. 43

Find the images p1 of radical center in the inversion relative to circlex d5, d6 and d4. Obtain, respectively, points p5, p6 and p7.

Fig. 44

Draw the circle d7 passing it through the points p5, p6 and p7.

Fig. 45

Find the center of the circles d6, d4, and d5 - poinst p8, p9 and p10.

Fig. 46

Construct triangle with sides o1, o2 and o3 based on points p8, p9 and p10.

Fig. 47

Find the center of the original circles - points p11, p12 and p13.

Fig. 48

Сonstruct triangle with sides o4, o5 and o6 based on points p11, p12 and p13 .

Fig. 49

Draw a circle d9, orthogonal to the original circle d1, d2 and d3. It should be noted that this circle passes through six specific points - the intersection point of a triangle with sides o4, o5 and o6 circles and d4, d5 and d6.

Fig. 50

Now we shall make a similar construction, but will assume that two of the three initial circles are imaginary radius.

Start a new project.

Let it be three actual circles defined on the plane. Single out two of them one by one by pressing the key on the keyboard with the Latin character i. This action will cause the actual radii of the circles that are converted to the corresponding imaginary.

Fig. 51

Construct the intersection point p1 and p2 actual circumference d1 and the imaginary circle d2. To do this select circle d1, then hold the Shift key down, select the mark non-visual circle d2 (a circle with a radius of imaginary and can be identified by specifying the cursor to a crosshair, indicating its actual center). After isolation, press the button with the Latin symbol p. Then highlighted the imaginary point, draw through them the real radical axis by pressing key on the keyboard with the Latin symbol o.

Fig. 52

Construct a line o2 similarly.

Fig. 53

Find the intersecting point p5 of lines o1 and o2.

Fig. 54

Construct the image p5 of point p6 in inversion relative to circle d1.

Fig. 55

Draw circle d4 passing through the point p5 perpendicular to the line o1 and the circle d1.

Fig. 56

Draw the circle d5 passing it through the point p5 perpendicular to the line o2 and the circle d1. As can be seen from the construction, the two circles intersect at a point p6.

Fig. 57

Circles d2 and d3 intersect in imaginary points p7 and p8.

Fig. 58

Passing line o3 through p7 and p8, we make sure that it passes through the point p5.

Fig. 59

Now we consider this case the location of the three circles, which you can not hold a valid orthogonal circle and, as a consequence, it is impossible to solve the problem of Apollonius, relying on the orthogonal property of real objects. Show that this problem can be easily solved with the use of previously considered inversion properties and obtain the imaginary circle orthogonal to three given circles.

Start a new project. Locate three circles d1, d2 and d3 on a plane, so that they clearly intersect eac other.

Fig. 60

Find the radical center of the circles - the point p1.

Fig. 61

Find images of radical center p1 in inversion relative to the circles d1, d2 and d3 - points p2, p3 and p4.

Fig. 62

Draw a circle d4 passing through obtained points p2, p3 and p4.

Fig. 63

Draw a circle through the radical center p1 and all possible pairs of a triad of points p2, p3 and p4. Obtain circles d5, d6 and d7.

Fig. 64

Determine the center of the original circle - points p5, p6 and p7.

Fig. 65

Construct a triangle with the sides o3, o4 and o5 based on points p5, p6 and p7 .

Fig. 66

Let us determine points p8, p9 o3 crossing the line o3, formed by the centers of the circles d1 and d2 - points p5 and p6 and the circle passing through the radical center and inverse points obtained as a result of the inversion of the center of the circles d1 and d2. Similarly constructed point p10 and p11 (from circles d1 and d3), and points p12 and p13 (circles of d2 and d3).

Fig. 67

Define the distance c1, c2, and c3 of the radical center point p1 to p9 (p8), p11 (p10), p13 (p12). Let us make sure that they are all equal to some imaginary value.

Fig. 68

Using the function of constructing a circle by specifying the center and radius define a circle d8 centered at p1 and radius c1.

Fig. 69