. Plane and solid analytic geometry; an elementary textbook. e 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 y - yv Substituting these values in (3), we have 4 x2 + 4 y2 — 4 xxx — 4 yxy + xx2 + yx2 = r2. But xx2 -j- yx2 = 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 OPi as a diameter, since its centre is at the point [ -i, ^and it passes through the origin. When, as in the above problem, we have to det
. Plane and solid analytic geometry; an elementary textbook. e 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 y - yv Substituting these values in (3), we have 4 x2 + 4 y2 — 4 xxx — 4 yxy + xx2 + yx2 = r2. But xx2 -j- yx2 = 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 OPi as a diameter, since its centre is at the point [ -i, ^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 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, x2 + y2 = r2, which pass through a fixed point, (xv y±), not on Cn. VIII, § 61] LOCI 97 Let jPj be a fixed point through which the secant PXP3passes, and let it be required to find the locus of P\ themiddle point of P^PZ- Transform the equation of thecircle to polar coordinates, with P1 as origin. The equa-tions of transformation are (by [20] and [24]), x z= x, + p cos 0, (1) y = yx + p sin 0, and the equation of the circle becomes (2) p2 + 2 (xx cos 0 + yx sin 0) p + xt2 + y? - r2 = p and 0 be the polar coordinates of P. The vec-torial angles of Pv P1, and P3 are evidently the same,and if 6 be substituted for 0 in (2), the solution of theresulting equation, p2 + 2 (>! cos 0 + ^ sin 0) /> + x2 + yx2 - r2 = 0, for p will give p2 and /o3, the two values of p for the pointsP2 and P3. But ,?2 + / r _ 2 - and />2 -f- />3 = — 2(2^ cos0r + ?/x sin0). (See Art. 8.) ence // = — (xx cos 0 + yx sin 0). This equation expresses the relation which must exist between the polar coor
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