CIRCLES AND OF SPHERES. 5 



In the above table, we have added the number of solutions of which each 

 problem is susceptible, or the number of different tangent circles which fulfil the 

 conditions. This number can, of course, be reduced in each case by particular 

 suppositions, but the number given above refers to the general problem. Par- 

 ticular suppositions can also be made in which some of the cases would become 

 impossible, or the circles become straight lines, or circles with infinite radii. 

 This table exhibits all the problems which occur in the tangencies of circles, the 

 geometrical elements being the point, the right line, and the circle — all being in 

 the same plane. 



The fact that the tangent circle is the limit of all secant circles, leads to the solu- 

 tions which I shall present, and which will be more fully explained in the solution 

 of' Problem 4. 



The solutions of Problems 1, 2, 3, 7, and 8 (as see Plate I), are very simple. In 

 Problem 2, extend the line A B, joining the two given points A and B, until it 

 meets the given line at 0. Draw any secant circle, as A B M N, through the two 

 given points. Let the radius and position of the centre of this secant circle be 

 changed until the two points of intersection, M and N, come together; then this 

 circle will become the required tangent circle. From 0, draw a tangent line, as 

 E, to this secant circle. Lay off the distance O E in each direction, as D and 

 O D\ and D and D are the points 'of contact of the two circles which will fulfil 

 the conditions. For, &$ A x B= 1>\ the circle drawn through the points 

 A, B, and D, will be tangent at D to the given line. The same reasoning applies 

 to the point D. 



In Problem 3, C being the given point, let fall a perpendicular to the line A L, 

 which is drawn to bisect the angle formed by the two given lines, and lay off 

 N D = N G. The point D will lie on the circumference of the recpiired circle, as 

 A L must be an indefinite diameter of that circle. The problem is thus reduced 

 to Problem 2, to draw a circle through the two points C and D, and tangent to the 

 line A E. 



In Problem 8, draw the auxiliary line L D parallel to L I, and distant from it 

 D F= A P, the radius of the given circle. From A, the centre of the given 

 circle, let fall A B perpendicular to L II (bisecting the angle), and lay off KB= A K. 

 Then if, through B and A, circles BAD and B A D be described tangent to L' D 

 (by Problem 2), and OD and HD be drawn, we have F and I, the points of contact 

 of two of the required circles. For D = O A = N, and FI) = PA = MN; there- 

 fore, F '= P = M, H I = H P'. By drawing an auxiliary parallel line L" S 

 within the angle, and distant also the radius of the circle, by a like process, two 

 more tangent circles, X Y and Q T, will be found to fulfil the conditions of the 

 problem. There are thus four solutions to the problem ; two of the circles convex, 

 and two concave to the given circle. 



&*• 



Problem 4. To draw a circle through two points, and tangent to a given circle. — 

 Draw through the two given points .Band A (see Fig. 1, Plate II), any circle, as 



