SECTION V.-Proportionals in the Circle.

Segments of Secants.—When two secants drawn in the plane of a circle intersect one another important relations exist between the segments of the secants determined by the circle. For instance,

(a) When two secants start from the same point, the segments of one form the extremes of a proportion of which the segments of the other form the means.

Let two secants from the same point E (Fig. 176) cut the circle in the points A, B, C, D; then

The proportion follows from the similarity of the triangles E A C, EDB.

Hence, whatever may be the direction of a secant from a point, E, in the plane of a circle, the rectangle between the segments is the same. If the point be upon the circumference, there are only two lines; and we may suppose one term in each ratio to become zero, or E B' x 0 = EB × 0.

B

C

A'

B'

FIG. 177.

E

Suppose now the secant E B', drawn from an exterior point E, to be turned about the point E until it coincides with E C, the tangent through E, then both the parts EA' and E B' become equal to E C, and therefore

(b). Hence, if from a point without the circle a tangent and a secant be drawn, the tangent is a mean proportional between the whole secant and the part without the circle.

Secants of Two Circles.-In any two circles, the secants joining the extremities of parallel radii drawn in opposite directions intersect at the same point in the line joining their centres.

Let O, O'be the centres, and O A, O' A' two parallel radii (Fig. 178). Join A A', then we have

OC:OA=O'C: O'A,

and therefore OC: OA

OC+O'C:OA + O'A'.

In this proportion the last three terms remain the same for all directions of the parallel radii, and therefore the line OC is constant; that is, it is not changed in length if OA be turned about O and O'A about O' so that they are always parallel.

In the same way we may show that if the circles are unequal, and O A, O'A' are parallel and directed the same way from the

FIG. 178.

centres, then A A' meets the line O O' produced in C'; and this will be

the case whatever be the inclina

tion of the radii. (Fig. 179.)

The converses of these theorems are also true, and are illustrated by the following mechanical contrivances for connecting two unequal wheels, so that they may revolve in the

FIG. 179.

O'

same time. Let O A, O'A' be two arms placed along parallel radii, and let the line A A' meet O O' in C.

If we can now bind the arms OA, O' A' together so that A A' is

always the same length and in the line A C, then OA and O'A' will remain parallel when the wheels revolve, and the time of rotation will be the same. For this end the arms O A, O A' are fixed to the rod C'A by two buttons which run in a groove. The rod is movable about the fixed axis C'. (Fig. 180.)

An analogous arrangement will give to two wheels an equal but

opposite rotation (Fig. 181), the points C being taken, so that OA: O'A' OC: O'C.

As the wheels revolve any point B in the connecting-rod moves

backwards and forwards on an arc bc, and conversely, an oscillating movement on bc will produce the rotation of the

wheels. Fig. 182 shows how the rectilineal movement of a shaft T may produce the rotation of the wheels O and O'.

Tangents Common to Two Circles.-A tangent common to two circles may be considered as a secant joining the extremities of two parallel radii, which are at the same time perpendicular to this secant; so that the tangents common to two circles pass through the fixed points C and C' determined above. If then we draw through the points C and C' tangents to one of the circles, they will at the same time be tangents to the other circle. (Fig. 183.)

The point C will be henceforth referred to as the point of intersection of exterior common tangents,

FIG. 183.

and the point C as the point of intersection of interior common tangents of the circles.

The Decagon and Pentagon.-In our treatment of the constructions of regular polygons the decagon and pentagon were omitted, because the readiest methods of constructing these figures depend upon principles of proportion.

The chord of an arc of 36° is the greater segment when the radius is divided in extreme and mean ratio. Hence we may make the following constructions :—

1. The Decagon.-Divide the radius in extreme and mean ratio, and take upon the circumference the greater segment A B (Fig. 184) ten times consecutively: the tenth point will coincide with A, and the lines joining the points of division will form a regular decagon; for each side will subtend an arc of 36°. 2. The Pentagon.-The same construction gives A C, the side of the regular inscribed pentagon.

FIG. 184.

B

This figure may be directly constructed in the following manner :Draw two diameters, A B, O C, perpendicular to

each other (Fig. 185), Take the middle, M, of

OA, and from M, as centre, describe the arc CD; OD will be the side of the decagon, CD that of the pentagon. It may be easily demonstrated that a right-angled triangle having for the sides containing the right angle the radius and the side of the decagon, has the side of the pentagon for hypothenuse.

C

60

3. The Quindecagon.—The angle at the centre of a quindecagon is 24°; that is, the difference between the angle 60° at the centre of a hexagon and the angle 36° at the centre of a decagon. Having described the circle (Fig. 186), mark off the chord, A B, equal to the radius, and from A mark off A C, the side of the decagon; the chord B C subtending the difference of the two arcs is the side of the quindecagon.

136

FIG. 186.

The triangle formed by joining the extremities of the side of a regular decagon with the centre, or the extremities of the side of a regular pentagon with the opposite angle has a vertical angle of 36°, and the angles at the base of 72° each, or just twice the vertical angle. Hence, to construct an isosceles triangle having the vertical angle half the angle at the base, we must draw a straight line, divide it in extreme and mean ratio, and construct a triangle having the

larger segment as base, and each of the two sides equal to the whole line.

The Auxiliary Curve.-When a regular polygon, not included amongst those to which the preceding constructions apply, has to be inscribed in a circle the side may be found by means of an auxiliary curve.

Upon a straight line take equidistant points (Fig. 187); number them in order, and at each point erect a perpendicular to this straight line. Upon perpendicular 3 take the side of an inscribed equilateral

triangle; upon 4, that of an inscribed square; upon 5, that of the pentagon; and so on with the divisions, 8, 10, 12, 15, 16, etc. Pass through these perpendiculars a curved line; the part intercepted upon the perpendicular 7 will be the side of the polyof seven sides inscribed in the same circle; the part intercepted upon No. II would be the side of a polygon of eleven sides, and so on.

3 4 5 6 7 8 9 10 11 12

gon

FIG. 187.

Arithmetical Questions.

1. Taking the radius of a circle as unity, find the length of the sides of the following inscribed figures :

2. Prove that if a be the side of a regular polygon inscribed in a circle the radius of which is unity, and a' that of a polygon in the same circle with twice as many sides, then

3. Use the formula in the last example to find the length of the side of the following inscribed figures, the radius being the unit :

Ist. Regular octagon.

2nd. Dodecagon.

3rd. Decagon.

4. Find the side of the following circumscribed figures, taking the radius as unity:

1st. Equilateral triangle.
2nd. Square.

3rd. Pentagon.

4th. Hexagon.

5. Prove that if b be the side of a regular polygon described about a circle the radius of which is unity, and b' that of a described polygon with twice as many sides, then