*73. Circles and conics as point-rows of the second order.* It is not difficult to see that a circle is a point-row of the second order. Indeed, take any point _S_ on the circle and draw four harmonic rays through it. They will cut the circle in four points, which will project to any other point of the curve in four harmonic rays; for, by the theorem concerning the angles inscribed in a circle, the angles involved in the second set of four lines are the same as those in the first set. If, moreover, we project the figure to any point in space, we shall get a cone, standing on a circular base, generated by two projective axial pencils which are the projections of the pencils at _S_ and _S’_. Cut across, now, by any plane, and we get a conic section which is thus exhibited as the locus of intersection of two projective pencils. It thus appears that a conic section is a point-row of the second order. It will later appear that a point-row of the second order is a conic section. In the future, therefore, we shall refer to a point-row of the second order as a conic. [Figure 14] FIG. 14 *74. Conic through five points.* Pascal’s theorem furnishes an elegant solution of the problem of drawing a conic through five given points. To construct a sixth point on the conic, draw through the point numbered 1 an arbitrary line (Fig. 14), and let the desired point 6 be the second point of intersection of this line with the conic. The point _L = 12-45_ is obtainable at once; also the point _N = 34-61_. But _L_ and _N_ determine Pascal’s line, and the intersection of 23 with 56 must be on this line. Intersect, then, the line _LN_ with 23 and obtain the point _M_. Join _M_ to 5 and intersect with 61 for the desired point 6. [Figure 15] FIG. 15 *75. Tangent to a conic.* If two points of Pascal’s hexagon approach coincidence, then the line joining them approaches as a limiting position the tangent line at that point. Pascal’s theorem thus affords a ready method of drawing the tangent line to a conic at a given point. If the conic is determined by the points 1, 2, 3, 4, 5 (Fig. 15), and it is desired to draw the tangent at the point 1, we may call that point 1, 6. The points _L_ and _M_ are obtained as usual, and the intersection of 34 with _LM_ gives _N_. Join _N_ to the point 1 for the desired tangent at that point. *76. Inscribed quadrangle.* Two pairs of vertices may coalesce, giving an inscribed quadrangle. Pascal’s theorem gives for this case the very important theorem _Two pairs of opposite sides of any quadrangle inscribed in a conic meet on a straight line, upon which line also intersect the two pairs of tangents at the opposite vertices._ [Figure 16] FIG. 16 [Figure 17] FIG. 17