perpendicular to it CX. Then evidently (Euc. III. 36) OA.OX OC. OZ=2; hence CX is the polar of A.

=

Thus as the point B moves along the line AB its polar

turns around, or envelopes, C. At Z therefore the polar is the chord of contact of tangents through that point to the

circle.

EXAMPLES.

1. The extremities of any diameter of a circle which cuts a given one orthogonally are conjugate points with respect to the latter.

2. If a variable chord AB of a circle pass through a fixed point P; the locus of the intersection of tangents to the circle at A and B is a line.

[The polar of P with respect to the circle.]

3. The diameter AB of a circle is the polar of a point at infinity in a direction perpendicular to AB.

4. The locus of a point which has a common conjugate with respect to three circles is their common orthogonal circle.

75. Theorem.-If A and B be any two points and L and M their polars with respect to a circle, the point LM is the pole of the line AB.

For LM is conjugate to both A and B, hence the line joining A and B is its polar (Art. 73), or "the line of connexion of any two points is the polar of the point of intersection of the polars of the points." Townsend.

76. More generally for three points A, B, C and their polars L, M, N, denoting the points MN, NL, LM by A', B', C' respectively; we see as above that A', B', C' are the poles of BC, CA, AB; hence, for any two triangles if the vertices of either are the poles of the corresponding sides of the other; then, reciprocally, the vertices of the latter are the poles of the corresponding sides of the former.

Def. Such triangles are said to be Reciprocal Polars with respect to the circle.

77. In the particular case when ABC and A'B'C' coincide, the triangle is Self-Reciprocal with respect to the circle. It is manifest, since each vertex is the pole of the opposite side, every two of its vertices are conjugate points; and the triangle is therefore termed Self-Conjugate with respect to the circle.

Its centre O coincides with the orthocentre O of ABC and the square of its radius (p) is given by

p2=OA.OX=OB. OY-OC. OZ, where X, Y, Z are the feet of the perpendiculars of the triangle.

This circle is called the Polar Circle of the triangle.

NOTE. In order that the polar circle may be real, the pairs of points A and X, B and Y, Cand Z, which are inverse with respect to it, must lie in the same direction from its centre 0. It is therefore

real when the triangle is obtuse angled, and imaginary for acute angled triangles.

78. Expressions for the Radius (p) of the Polar Circle.

Let O be the ortho-centre of ABC, then it appears that A, B, C are the ortho-centres of BOC, COA, and AOB respectively. For this reason the four points A, B, C, O are said to form an Orthocentric System.

Also the circum-circles of the four triangles BOC, COA, AOB, and ABC are equal.

Hence since a and AO, chords of equal circles, subtend complementary angles at the circumferences,

also (fig. B)

a2+AO2=b2+BO2=c2+ CO2=d2,..

therefore by substitution from (1)

or

.(1)

(Euc. II. 13)

(2)

a2=2d2-b2-c2+200. OZ, -CO.OZ=d2-1(a2+b2+c2)= p2........

2

This formula is equivalent to p2=d2cos A cos B cos C, by reduction or independently, as follows:

-p2=OC. OZ=0C.

OA.OB
d

=

= d'cos A cos B cos C,...(3)

since a chord is equal to the diameter of the circle into the sine of the angle it subtends.

EXAMPLES.

1. The four polar circles of the triangles BOC, COA, AOB and ABC are mutually orthogonal.*

[Let their radii be Pa, Pb, Pc, p. Since their centres are at A, B, C, O, by Euc. II. 2,

therefore, etc.]

AB2=AB. AZ+AB. BZ=på2+po2;

2. B and C, C and A, A and B are pairs of conjugate points with respect to the polar circles of BOC, COA and AOB respectively.

3. The square of the distance BC between any two conjugate points is equal to the sum of the squares of the tangents drawn from them to the circle.

[By Ex. 1 the tangents from B and C to the circle pa are the radii of Рь and but BC2= p2 + på2; therefore, etc.]

4. Prove that AZ. BZ=ť2, where t is the tangent to the polar circle from Z, the Polar Centre † of AB; and conversely.

[By similar triangles ACZ and OBZ, AC: CZ=OZ: BZ, etc.]

5. Conjugate points A and B with respect to any chord MN are conjugate with respect to the circle.

[For the polar centre Z of AB is the middle point of MN; but (hyp.) ZA. ZB=ZM2=-ZM.ZN or the square of the imaginary tangent from Z to the circle; therefore, etc., by Ex. 4.]

6. If a number of circles have a common orthogonal circle, the extremities of any diameter of the latter are conjugates with respect to the entire system.

7. On a given line find two points which shall be conjugates to each of two given circles.

[The middle point of the required segment is such that the tangents from it to the circles are equal; therefore, etc., by Art. 72, Ex. 3.]

* Hence :-If four circles are mutually orthogonal, their centres form an orthocentric system and one of the circles is imaginary.

+ Z the foot of the perpendicular from the centre on AB is also called the Middle Point of the line. (Cf. Euc. III. 3.)

8. On a given circle O find two points A and B which shall be conjugates to each of the circles C, r1; D, r2.

[The middle point M of the required chord is on the radical axis L of the given circles (Art. 72, Ex. 3). Let 2t be the length of AB; then CM2=t2+r2=r12+AM2=r12+r2-OM2;

hence CM2+OM2 is known, and the triangle COM is completely determined; therefore, etc.]

9. Place a chord of given length in a circle so that its extremities may be conjugates with respect to another.

[See Ex. 8.]

10. If a right line AB meet either (C, r) of two circles in conjugate points (A, B) with respect to the other; then reciprocally it meets the latter (C', r') in conjugate points (A' and B') with respect to the former.

[For by Ex. 5 AB divides A'B' harmonically, hence A'B' divides AB harmonically; therefore, etc.]

11. Find the locus of the middle points M and N of the chords AB and A'B' in Ex. 10.

[CM2+C'M2 CN2+C"N2=CM2+ C'N2+MN2

=CM2+C'N2+MB2+A'N2

=2+2=const. ;

(Art. 71.)

hence the required locus is a circle whose centre 0 is at the middle point of CC' and the square of whose radius is equal to (2+2)-83,