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{x-(a, COS X-C)} 2+{y y(a, sin X+b)} 2 =b2+C2-d2 (9) Here, (b2+c2-d2) is a constant. Consequently, equation 8 defines a circle with center at (a, cos X-c)+j(al sin X +b) and a radius given by Vb2+C2 -d2 Such an impedance circle is shown in Fig. 4, where 0 is the origin, and D the center of the circle.The operating point corresponding to a particular value of 0 may be determined as follows: Determine the point C such that the vector OC is given by aleix. Draw a straight line E2'CE2 through the point C, making an angle 0 with the positive direction of j as shown in Fig. 4, the angle o being reckoned positive if CE2 leads the positive j direction in the counterclockwise direction. The vector OE2 then gives the operating point corresponding to aiejX+j{(b cos 0+c sin 0)+ V,(b cos 0+c sin 0)2 -d2}e3O while the vector OE2' corresponds to aiej -+j{(b cos 0+c sin 0)-V(b cos 0+c sin 0)2 -.d2}e'G In Figs. 5 and 6, current and conjugate impedance loci for the stator and the rotor respectively of the test machine are shown. In both these figures operating points for a few values of a are shown; the point C' being the conjugate of the point C previously mentioned.

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