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 refer

ence to both circles is the

same (Prop. 22),

PA2 - r2 = PB2 — No12,

-

Hence, O is a fixed point, and the radical axis is perpendicular to AB through it.

Corollary 1.-If two circles intersect or touch, their common chord or tangent is a radical axis.

Corollary 2.-The tangents drawn from any point in the radical axis to the two circles are equal to each other.

Hence, if TT" be a common tangent, CT CT"; and the radical axis may be found by bisecting two common tangents and joining the points of bisection.

Corollary 3.-The radical axis of two equal circles passes through the middle of the line through their centres.

Proposition 24.

Theorem.—The radical axes of three circles, taken two at a

time, pass through the same point.

Let A, B and C be three circles; the radical axes of A and B and of B and C will meet in some point P.

Now, because P is in the radical axis of A and B, the powers of P with reference to A and B are equal; hence, also, the powers of P with reference to B and C are equal; therefore the powers of P with reference

BL

to A and C are equal, and P lies in the radical axis of A and C. Hence, the three radical axes intersect in the point P..

Definition. The point P is called the radical centre of the three circles.

Scholium 1.—If the centres of A, B, C be in a straight line, the radical centre is at an infinite distance.

Scholium 2.-This affords an easy method of finding the radical axis of two circles. Draw a third circle cutting the two, and find the intersection of two common chords. This will be a point (Prop. 23, Cor. 1) in the radical axis, which may be drawn perpendicular to the line through the centres.

SECTION VIII.

CENTRES OF SIMILITUDE.

DEFINITIONS.

1. THE centres of similitude of two circles are the points in which the line join

ing their centres is divided in the ratio of the radii.

Thus, if R and R' represent the radii of

the circles A and B, if

AC BC: AC': BC' :: R: R',

B

then C is the internal, and C' the external centre of similitude.

--

Corollary. If the two circles touch each other, the internal centre of similitude is in the point of contact.

Proposition 25.

Theorem.-A line through the extremities of two parallel radii of two circles passes through a centre of similitude.

Let A and B be two circles, and AD, BF and AD, BE be parallel radii; if DF, DE be joined, they will cut AB in C and C", the internal and external centres of similitude.

For

and

AC: BC: AD: BF :: R: R',

AC: BC' :: AD: BE :: R: R'.

Hence, C and C' are the centres of similitude.

C'

Corollary 1.-Any transversal through the centre of similitude is divided in the ratio of the radii.

Corollary 2.-Any tangent through the centre of similitude is divided in the ratio of the radii.

Corollary 3.-The line joining the centrès is divided harmonically at the centres of similitude.

Definitions.—The points E and D, as also M and N, are said to be homologous with respect to each other; and E and N, as also M and D, are said to be anti-homologous with respect to each other. If C'N, C'n be two secants, then Em, Nn are said to be anti-homologous chords.

Proposition 26.

Theorem.—The product of the distances of a centre of similitude of two circles, from two anti-homologous points, is constant.

or C'D. C'N = P. Also (Prop. 25, Cor. 1),

Dividing the first by the second, C'D. C'M

Now, D and M are anti-homologous points, and the second member of the equation is constant. Hence the theorem.

Corollary 1.-The anti-homologous chords meet on the radical axis.

Let Em, Nn be anti-homologous chords meeting at O. Then, from the above,

C'E. C'N- C'm. C'n;

hence (V. 27, Cor. 3), E, m, n and N are on the circumference of the same circle, and Em is a common chord of this new circle and the circle B; hence (Prop. 23, Cor. 1) it is a radical axis of these two circles. For the same reason Nn is a radical axis of the new circle and the circle A. Hence, O is the radical centre of the three circles, and OP (Prop. 24) the radical axis of A and B.

Corollary 2.-The extreme positions of the chords (that is, when they become tangents) also intersect on the radical axis.

Proposition 27.

Theorem.-If three circles be given, and considered, taken two at a time, as forming three pairs—

1. The external centres of similitude of the three pairs are in a straight line.

2. The external centre of similitude of any pair, and the internal centre of similitude of the other two pairs, are in a straight line.

Let A, B and C be three circles, C', C", C"", their external