6 THE TANGENCIES OF 



B A CD, secant to the given circle M N D, and cutting it in the two points Cand D. 

 If the radius ami position of the centre of this circle he changed until the two points of 

 intersection G and D come together, it icill become the required tangent circle. Join 

 1) G, and extend the line until it meets B A prolonged at Z. From Z, draw two 

 lines, ZM&nd Z N, tangent to the given circle; the points of contact iVand M will 

 be the points of contact of the required tangent circles. For Z A x Z B= Z C 

 x ZD = Z N- = Z IP; therefore, the circle B A JVwill be tangent at N to the 

 line Z N, and both circles, being tangent to the same right line at the same point 

 iV^ must be tangent to each other. The same course of reasoning applies to the 

 other tangent circle BAM. There are, then, two solutions ; one tangent circle 

 convex, the other concave to the given circle. 



With the use of any other secant circle, as B A L P, by a like construction, the 

 same tangent circle B A iV would be obtained, and the same point of contact N. 

 Therefore, the secant line drawn through PandL will pass through the same point 

 Z as in the case of the secant line D G. Fig. 2 exhibits the case when the two 

 given points are within the given circle. 



Therefore, it follows that if, through any two points in the plane of a given 

 circle, any number of secant circles be drawn intersecting said circle, and right 

 lines be drawn through the points of intersection, they will all meet in a common 

 point on the prolongation of the line joining the two first named given points. 



This solution of Problem 4 is derived, it will be seen, from the fact that the 

 tangent cikcle is the limit of all secant circles. The tangent circle B A N is 

 the limit of all secant circles, as B A G D, B A L P, &c. This principle, and the 

 construction in this problem, will be found of universal application in all that follows. 



Problem 5. To draw a circle through a given point, and tangent to a given right 

 line, and to a given circle. — Draw the line G B (see Fig. 1, Plate III) through the 

 centre of the given circle, and perpendicular to the given right line C E. Suppose 

 Pis the given point, and, for the purpose of analysis, that P Eis the required 

 tangent circle. S D, joining the centres, will pass through the point of contact 0. 

 Join B and E. B E will be one continuous straight line. For, D E and 

 B /S being parallel, the angles 8 B and ODE are equal, and, the triangles being 

 isosceles, the angles at are equal. In the similar triangles B A and B G E, 

 we have B C x B A = B E x BO. Join B P, and suppose that P' is the point in 

 which B P cuts the required circle. BPxBP'=BExBO. Therefore, B P 

 x B P'= B Ox B A. But B G, B A, and B P are given distances; therefore, the 

 point P can be found by taking B P' equal to the fourth proportional to B P, B G, 

 and B A. This will readily be done by making the angle B AP' = the given angle 

 B P G ; or, passing a circle through the three given points C, A, and P, it will cut 

 B P in the required point P. Then, by Problem 4, through the two points P and 

 P', draw a circle tangent to the given circle; it will also be tangent to the given 

 right line. This gives two circles, PP' 0, PP O, fulfilling the required conditions. 



Also join PA. By a similar process, we shall find the triangle A E' G similar 

 to the triangle A O" B, and A Px A P" = A E' x A 0'" = A B x A G. Find, 

 then, A P" a fourth proportional to A P, A B, and A G. This can be done by 



