y TIJE TANGENCIES OP 



to both of the given circles, one concave to both, one convex to F and concave to E, 

 and the fourth concave to F and convex to E. 



We shall hereafter name the point the "external similar point," and the point 

 O the "internal similar point' of the two circles. 1 



We will now suppose that one of the given circles encloses the other, as in Fig. 3. 

 If the extremities of any two parallel radii E Q and F C, on opposite sides of E F, 

 are joined, the line C Q will cut the line joining the centres at a fixed point 0. 

 If the extremities of parallel radii, as F N, and E Q, on the same side of the line 

 E F, be joined by the line Q N, this line, prolonged, will meet the line joining the 

 centres at a fixed point O. And it will readily be seen that they are the "similar 

 points" of these circles, and that, by using these points and 0' in the same man- 

 ner as and 0' in the former cases, the four solutions of the problem (exhibited 

 to the eye in Fig. 3) will be obtained. 



It is worthy of note, that in Problem 5 (see Fig. 1, Plate III) we can regard the 

 points B and A as every way analogous to the points and 0' in Problem 6. Re- 

 garding the given right line C E as the circumference of a circle of infinite radius, 

 B G is the line joining the centres of the two given circles, and B and A are the 

 points where the lines B i^and A M, tangent to both (and parallel to G E), inter- 

 sect the line joining the centres. The points B and A answer precisely, therefore, 

 to the description of the "external and internal similar points" O and O' in 

 Problem 6. 



Problem 9. To draw a circle tangent to a given rigid line and to two given circles. — 

 In this problem, as see Fig. 1, Plate V, draw the auxiliary line KB parallel to the 

 given line, G G, and at a distance equal to the radius of the smaller of the two 

 given circles, and the auxiliary circle B N with a radius equal to the difference of 

 the radii of the circles. Then, by Problem 5, through the point F, draw a circle, 

 as F B D, tangent to the circle B N, and to the auxiliary line K D. The centre 

 of this circle will be the centre of the required circle. For, join O, the centre of 

 this tangent circle, with A and F, and let fall O D perpendicular to the given line, 

 we shall have E F= G D — B S; and, taking these distances from the radius of 

 the circle F B D, we have O S= O E= O G. Therefore, a circle passed through 

 S, E, and G will be tangent to the two given circles at S and E, and to the given 

 right line at G So, by the use of a similar auxiliary line, T P, on the other side 

 of the given line, the circle N P F is found, whose centre, L, is the centre of 

 another circle, MR E', which will fulfil the conditions of the problem. 



Thus, in each case, the centre of the required circle is found by obtaining the 

 centre of an auxiliary circle, whose radius differs from the required one (either less 

 or greater) by the radius of the smallest of the given circles. 



It will be found that there are eight solutions to this problem. They are all 



- 



1 These are the names given to these points in the dissertations of the German mathematician Steiner, 

 in the first volume of Crelle. It is proper to add that the writer had made use of said points, and, 

 indeed, arrived at nearly all these solutions, in 1838, and communicated them to his friends at West 

 Point, long before he had seen the very curious German memoir referred to. 



