146 UNIVERSITY OF COLORADO STUDIES 



and of the v-axis 



X — o. 



The equations of the inner and outer bi-sector are therefore 



x (k — 1 1 + k 2 ) + y — ak — b = o, 



and 



x (k + V 1 + k 2 ) + y — ak — b = o. 



As is easily verified, the co-ordinates (25) which represent the center 

 of one of the circles of the locus satisfy the first of these equations and 

 those of (20) the second. 



If we now use the technical terminology, i. e., designate the arcs 

 MB, OB, MA, AO, etc., as arcs of compound curves and their points of 

 tangency as points of compound curves, we may state the theorem: 



The locus 0} all points of compound curves between two given tangents 

 and points 0} curve consists 0) two circles which pass through the two 

 given points and whose centers lie on the bisectors oj the two given 

 tangents. 



To construct these centers we may, therefore, connect A with B, 

 erect a perpendicular to AB in the middle of AB, which will intersect 

 the bi- sectors in the required points P and Q. 



Any point of compound curve as B lies on the same right line with 

 the centers of the corresponding arcs of compound curves OB and MB. 

 Since O and B lie also on the circle of the locus of points of compound 

 curves with the center P, the perpendicular to the chord OB through 

 E passes through P. Hence 



z PEX = z PEG. 



This means that every ray connecting the centers of two compound 

 curves whose point of tangency, or point of compound curve, lies con- 

 stantly on one of the circles of the locus, is tangent to a fixed circle which 

 is concentric with the circle of the locus. The same can be proved for 

 the ray EF. The two concentric circles, one with P, the other with Q 

 as a center, in Fig. 3, are designated by (P) and (Q). 



The circle of the locus with P as a center intersects the ;y-axis and 

 the line V, in two other points J and H, so that JM = OH, and TM 

 — TO = OH. Evidently OH is equal to the diameter of the circle 



