Suppose S = 0, S′ = 0 are lines (that is, let S, S′ now denote linear functions Ax + By + C, A′x + B′y + C′), then S − kS′ = 0 (k an arbitrary constant) is the equation of any line passing through the point of intersection of the two given lines. Such a line may be made to pass through any given point, say the point (x0, y0); if S0, S′0 are what S, S′ respectively become on writing for (x, y) the values (x0, y0), then the value of k is k = S0 ÷ S′0. The equation in fact is SS′0 − S0S′ = 0; and starting from this equation we at once verify it a posteriori; the equation is a linear equation satisfied by the values of (x, y) which make S = 0, S′ = 0; and satisfied also by the values (x0, y0); and it is thus the equation of the line in question.
If, as before, S = 0, S′ = 0 represent circles, then (k being arbitrary) S − kS′ = 0 is the equation of any circle passing through the two points of intersection of the two circles; and to make this pass through a given point (x0, y0) we have again k = S0 ÷ S′0. In the particular case k = 1, the circle becomes the common chord (more accurately it becomes the common chord together with the line infinity; see § 23 below).
If S denote the general quadric function,
S = ax2 + 2hxy + by2 + 2fy + 2gx + c,
then the equation S = 0 represents a conic; assuming this, then, if S′ = 0 represents another conic, the equation S − kS′ = 0 represents any conic through the four points of intersection of the two conics.
| Fig. 52. |
14. The object still being to illustrate the mode of working with coordinates for descriptive purposes, we consider the theorem of the polar of a point in regard to a circle. Given a circle and a point O (fig. 52), we draw through O any two lines meeting the circle in the points A, A′ and B, B′ respectively, and then taking Q as the intersection of the lines AB′ and A′B, the theorem is that the locus of the point Q is a right line depending only upon O and the circle, but independent of the particular lines OAA′ and OBB′.
Taking O as the origin, and for the axes any two lines through O at right angles to each other, the equation of the circle will be
x2 + y2 + 2Ax + 2By + C = 0;
and if the equation of the line OAA′ is taken to be y = mx, then the points A, A′ are found as the intersections of the straight line with the circle; or to determine x we have