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.




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.





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.


1. If the point be external,

AP.PB = (V. 27)

[blocks in formation]


2. If the point be internal, EP.PF (V. 26) GP. PH



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.

Because the power of

any point P with reference to both circles is the same (Prop. 22),


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.

PA2 – r2 = PB2 — No12,


Proposition 23.

Theorem.—The radical axis of two circles is perpendicular to the line joining their centres.

if r, r' be the radii; or, PB-PA'2 — p2.


Now, draw PO perpendicular to AB, and

(I. 42),


PB2 – PA2 = OB2 – OA2 = p22 — p2.


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

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.




1. THE centres of similitude of two circles are the points in which the line joining their centres is divided in the ratio of the radii.


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

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

Thus, if R and R represent the radii of the circles A and B, if


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.


AC: BC: AD : BF :: R: R', AC: BC' :: AD: BE :: R: R'. Hence, C and C' are the centres of similitude.


In m


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

Let C' be a centre. of similitude of the circles A and B; then C'E. C'N is constant.

Let the power of C' with reference to the


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.


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