CIRCLES AND OF SPHERES. 7 



drawing the line B P" so as to make the angle A B P" = the angle A P G ; or, 

 pass a circle through P, B, and G, and it will cut PA prolonged at P". Then, by 

 Problem 4, through Pand P", draw circles 0'" P" E' and P" P E" tangent to the 

 given circle ; they will also fulfil the conditions of the problem. 



This makes four solutions in all ; two of the tangent circles convex, and two con- 

 cave to the given circle. 



Fig. 2 exhibits the case in which the given right line intersects the given circle, 

 the given point being within the circle. There are only two solutions in this 

 case, each circle being convex to the given circle. 



Problem 6. To draw a circle through a given point, and tangent to two given circles. 

 — In Fig. 1, Plate IV, let E and F be the two given circles, and A the given point. 

 Find the point 0, on the line joining their centres, where the line T T', tangent to 

 both on the same side of E F, will cut it. Suppose, for the sake of analysis. AG D 

 to be the required circle. Join the points of contact G and D. This line, pro- 

 longed, will pass through the point 0. For, mark the point L where this line cuts 

 the circle F. The triangles G F L, D D' G, and MD Pwill be similar. There- 

 fore, the angles MD i?and G L Fare equal, and the radius F L is parallel to the 

 radius ED. Therefore, the line D C L passes through the point 0, which is the 

 intersection of all right lines passing through the extremities of parallel radii on 

 the same side of E F. (See Biot's Analytical Geometry, Chap. II.) 



Join with the given point A, and suppose A' to be the point in which this line 

 cuts the required circle. The triangle C Sh similar to the triangle M N; but 

 M Nis also similar to D K. Therefore, the triangle C Sis similar to the 

 triangle D K, and K x S= D x G= Ax A'. Therefore, if we 

 lay off A' equal to a fourth proportional to A, K, and S, we obtain the 

 point A'. This point can be obtained by making the angle S A' = the angle 

 OAK. Or, pass through the given points A, K, and S, a circle, it will cut A in 

 the point A'. Then, by Problem 4, drawing, through A and A', the circles A A' G 

 and A A' C' tangent to one of the given circles, they will also be tangent to the 

 other. This construction gives two of the required circles. 



If, in Fig. 2, the line T T' be drawn tangent to the two given circles, the points 

 of contact being on different sides of the line joining their centres, the point 0', 

 in which these lines intersect, enjoys properties entirely similar to those found for 

 the point in Fig. 1. We shall find the triangles O G C", O MN, and 9 D K, 

 all similar, and O Kx O G = 0' D x O C" = O A x Q A". Thus, the triangle 

 0' G A" is similar to the triangle 0' A K; and laying off O A" equal to a fourth 

 proportional to O A, O K, and O G, we obtain the point A". 



A" is easily found in construction by drawing G A" so as to make the angle 0' GA'' 

 equal to the angle OAK. Or, passing a circle through A, K, and G, it also should 

 cut the line A O at A". Then, through A and A", by Problem 4, pass two circles, 

 A C" A" and A G'" A", tangent to the circle F, they will also be tangent to the 

 circle E. 



This completes the four solutions of this problem ; one of the circles being convex 



