Plane and solid analytic geometry; an elementary textbook . It is therefore possible !»»; ANALYTIC GEOMETRY [Ch. VIII, § 61 to eliminate the parameters and obtain a single equationin terms of the variables and constants only. Solving (1)and (2) for x2 and yv we have x2 = 2 x - xv and y2 = 2 yf -, yvSubstituting these values in (3), we have 4 a/» + 4 y2 - 4 a^ - 4 yty + zx2 + yx2 = r2. But a^2 + j/j2 = r2, and, dropping primes, the equation reduces to x2-{-y2-x1x-y1y = 0. This is the equation of the locus of P. It is a circle on 0P1 as a diameter, since its centre is at the point I -J, 2l J,and
Plane and solid analytic geometry; an elementary textbook . It is therefore possible !»»; ANALYTIC GEOMETRY [Ch. VIII, § 61 to eliminate the parameters and obtain a single equationin terms of the variables and constants only. Solving (1)and (2) for x2 and yv we have x2 = 2 x - xv and y2 = 2 yf -, yvSubstituting these values in (3), we have 4 a/» + 4 y2 - 4 a^ - 4 yty + zx2 + yx2 = r2. But a^2 + j/j2 = r2, and, dropping primes, the equation reduces to x2-{-y2-x1x-y1y = 0. This is the equation of the locus of P. It is a circle on 0P1 as a diameter, since its centre is at the point I -J, 2l J,and it passes through the origin. When, as in the above problem, we have to determinethe locus of a point situated on a moving line whichrevolves about some fixed point in it, polar coordinatesare often convenient. The fixed point is taken as thepole, and the distance from it to any position of the Y^ moving point becomes A the radius vector. The following prob-lem will illustrate themethod: 28. Find the locus ofthe middle points ofchords of the circle,. ;r r\ which pass through a fixed point, (xv t^), not on thecircle. Ch. VIII, §61] LOCI 97 Le1 J\ be a fixed point through which the secant PXPSpasses, and let it be required to find the locus of _P, themiddle point of P2Ps- Transform the equation of thecircle to polar coordinates, with P1 as origin. The equa-tions of transformation are (by [20] and [24]), x = X-, 4- p cos 6,(1) 1 y = Vi + p sin 0, and the equation of the circle becomes (2) p2 + 2 (xx cos 6 + yx sin 6) p + xx2 + y2 - r2 = 0. Let p and 6 be the polar coordinates of P. The vec-torial angles of P2, P, and P3 are evidently the same,and if 6 be substituted for 6 in (2), the solution of theresulting equation, p2 + 2 Oj cos 0 + sin 6) p + ^2 + yx2 - r2 = 0, for ^ will give p2 and p3, the two values of p for the pointsP2 and P3. But / = ^, and p., + p3 = — 2 (a^ cos 0r + ^ sin 0). (See Art. 8.) ence p = — (xx cos 6 + y1 sin0). This equation expresses the relat
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