Modern geometry . fig. 54. Suppose we have the radical axis and one circle of thesystem. Prom N (which must be outside all the circles) draw atangent NQ to the circle. With centre N and radius NQ describe a circle. Draw BR a tangent to this circle from any suitable pointin AN (or that line produced). Then the circle with B as centreand BR as radius will be a circle of the system. For AN-AQ=NQ=NR^=BN2-BR2. It should be noticed that instead of taking N as centre we might takeany point on the radical axis. This method would then apply to inter-secting circles as well as non-intersecting, Ex. 403.
Modern geometry . fig. 54. Suppose we have the radical axis and one circle of thesystem. Prom N (which must be outside all the circles) draw atangent NQ to the circle. With centre N and radius NQ describe a circle. Draw BR a tangent to this circle from any suitable pointin AN (or that line produced). Then the circle with B as centreand BR as radius will be a circle of the system. For AN-AQ=NQ=NR^=BN2-BR2. It should be noticed that instead of taking N as centre we might takeany point on the radical axis. This method would then apply to inter-secting circles as well as non-intersecting, Ex. 403. Draw a system of coaxal circles, one circle of the systemhaving its centre 4 cm. from the radical axis and having a radius of 3 cm. It is worthy of special notice that in a system of coaxalcircles one member of the system consists of the radical axis andthe line at infinity. Ex. 404. In fig. 54 what position of R will give the radical axis as amember of the system ? COAXAL CIRCLES.
Size: 1222px × 2044px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookauth, bookcentury1900, bookdecade1900, bookidcu31924001083090
