Corollary 1.—Because CPD is a right angle, and PC bisects the angle APB, PD must also bisect the angle BPF. Corollary 2.-If the angles APB, BPF be bisected, then AB is divided harmonically in C and D; and, conversely, if AB be divided harmonically in C and D, the angles APB, BPF will be bisected by PC and PD. SECTION VI. POLE AND POLAR IN THE CIRCLE. DEFINITIONS. IF from a point, O, without the circle ABD, the secant OAB be drawn, and C, the harmonic conju- B о D Proposition 20. Theorem.—The polar of a point is a straight line at right angles to the diameter through the point. Let O and C be harmonic conjugates with respect to A and B on the circumference ABD. Draw CF at right angles to the diameter DE; and on OC describe a semicircle passing through F. Join BF, and let it cut the circle in G; join Cor. 2) harmonically in D and E, and F is the harmonic conjugate of O. Hence F is a fixed point, and C is always in the perpendicular to OE through F; therefore CF is the polar of the pole O. Corollary 1.—The radius of the circle is a mean proportional between the distances of the centre from the pole and the polar. For, because as in last Prop., or, OE OD :: FE: FD, OK: DK:: DK: FK, OK.FK DK2 = R2. = Corollary 2.-The pole and polar are on the same side of the centre; if the pole be exterior, the polar is interior to the circle, and vice versa. Corollary 3.-The polar passes through the point of contact of the tangent through the pole. Corollary 4.-The polar of the centre is at infinity, and the polar of a point at an infinite distance passes through the centre. Corollary 5.-The polar of a point on the circle is a tangent through this point. Proposition 21. Theorem.-The polars of all points of a straight line pass through the pole of that line; and the poles of all straight lines which pass through a fixed point are in the polar of that point. CD. From the similar triangles DCE, FCG, CE. CGCD. CF. G Hence CD. CF = CH2 and FG is (Prop. 20, Cor. 1) the polar of D. Hence the polar of any point, D, of the line AB passes through G, and the pole, G, of any line, AB, which passes through D is in the polar FG of the point D. Scholium. From this it results that if the angles of the polygon ABCDE be the poles Each of the sides of the polygon MNOPQ, the angles of the second polygon are the poles of the sides of the first. polygon is said to be the reciprocal polar of the other with respect to the circle, which is called the directing circle. M E N If one polygon be inscribed in the circle, the other will be circumscribed about it, the sides of the latter passing through the angles of the former. The angles of either may be determined from the sides of the other, or the sides of either from the angles of the other. A double set of theorems may be deduced, in which the properties relating to the lines of one may be converted into properties relating to the points of the other, and vice versa. These theorems belong principally to curves of which Elementary Geometry does not treat, and will not be considered here. SECTION VII. RADICAL AXES. DEFINITIONS. 1. THE power of a point with reference to a circle is the rectangle of the segments of a secant through the point, formed by the circumference of the circle. Thus the power of P with reference to ABCD is PA. PB; of P', is PC. P'D. 2. Two circles cut each other orthogonally when the radii to the point of intersection are at right angles to each other. That is, when ACB is a right angle. B B D P A 3. The radical axis of two circles is the locus of all points whose powers, with reference to the circles, are equal. Proposition 22. Theorem.-The power of a point with reference to a circle is equal to the difference of the square of the distance of the point from the centre of the circle, and the square of the radius. If the point be on the circumference, its power is zero. Corollary 1.-If the point be external, its power is equal to the square of the tangent from it to the circumference. Corollary 2.-If two circles cut each other orthogonally, the square of either radius is equal to the power of its centre with reference to the other circle. Proposition 23. Theorem.-The radical axis of two circles is perpendicular to the line joining their centres. Because the power of any point P with reference to both circles is the same (Prop. 22), PA2-PB1 — 72, if r, r' be the radii; or, PB2 - PA2 = p22 — p2. Now, draw PO perpendicular to AB, and (I. 42), PB-PA OB2 - OA2 = r2 — go?. = B |