443. - Circles: Diameter and Perpendicular: Mean Proportional. - Let ABC (Fig. 294) be a semicircle. From C, any point in the curve, draw a line to A and another to B; then ABC will be a right-angled triangle (Art. 352). Draw the line CD perpendicular to the diameter A B; then CD will divide the triangle A B C into two triangles, A C D and CBD, which are homologous. For, let the triangle CBD be revolved on D as a centre until its line CD shall come to the position E D, and the line DB occupy the position D F, each in a position at right angles to its former position, the point B describing the curve B F, and the point C the curve CE, and each forming a "quadrant or angle of ninety degrees. Since these points have revolved ninety degrees, therefore the three lines of the triangle C BD have revolved into a position at right angles to that which they before occupied; hence the line EF is at right angles to CB, and (from the fact that A CB is a right angle) parallel with A C Since the triangle E FD equals the triangle CB D. and since the lines of E FD are parallel respectively to the corresponding lines of A CD, therefore the trianglesA CD and CB D are homologous.

Fig 294.

Fig 294.

Comparing the lines of these triangles and putting a - A B, y = CD, and x = D B, we have -

DB: D C:: DC: AD, x: y:: y: a - x, y2 == x (a - x). (160.)

Or, in a semicircle, a perpendicular to the diameter terminated by the diameter and the curve is a geometric mean, or mean proportional, between the two parts into which the perpendicular divides the diameter.

444. - Circle: Radius from Given Chord and Versed Sine. - Let A B (Fig. 295) be a given chord line and CD a, versed sine. Extend CD to the opposite side of the circle; it will pass through F, the centre. Join A and C, also E and

Section XI The Circle 405

Fig. 295.

B. The line A D, perpendicular to the diameter C E, is a mean proportional. between the two parts CD and DE (Art. 443); or, putting a = A D, b = C D, and r equal the radius F E, we have -

Cd: Ad:: Ad: De;

b: a:: a: 2 r - b,

a2 = b (2 r - b),

a2 = 2rb-b2,

a2 + b2 = 2rb,

r =a2+b2/2b (161.)

Or: The radius of a circle equals the sum of the squares of half the chord and the versed sine, divided by twice the 'versed sine.

Another expression for the radius may be obtained; for the two triangles CB D and CE B (Fig. 295) are homologous (Art. 443) and their corresponding lines in proportion. Putting f for CB, we have -

Cd: Cb:: Cb: Ce,

or - v: f:: f: 2 r,

or - f 2 = 2 r v,

and -

r = f2/2v. (162.)

Or: The radius of a circle equals the square of the chord of half the arc divided by twice the versed sine.

445. - Circle: Segment from Ordinates. - When the curve of a segment of a circle is required for which the radius cannot be used, either by reason of its extreme length, or because the centre of the circle is inaccessible, it is desirable to obtain the curve without the use of the radius. This may be done by calculating ordinates, a rule for which will now be developed.

Section XI The Circle 406

Fig. 296.

Let DCB (Fig. 296) be a right angle, and A DB a circular arc described from C as a centre, with the radius B C= CD = CP. Draw PM parallel with D C, and A G parallel with CB. Now, in the segment A D G, we have given A G, its chord, and D E, its versed sine, and it is required to find an expression by which its ordinates, as PF, may be computed. From Art. 416, we have -

Section XI The Circle 407

or, putting for these lines their usual symbols -

y2 = r2 - x2,

Section XI The Circle 408

now we have -

EC = DC-DE,

EC= FM,

FM=DC-DE,

FM =r-b.

Then we have -

PF=:PM-FM;

or, putting t for PF and substituting for PM and FM their values as above, we have -

t = y-(r-b),

and for y, substituting its value as above, we have -

(163.)

(163.)

Or: The ordinate in the segment equals the square root of the difference of the squares of the radius and the abscissa minus the difference of the radius and the versed sine.

For example: let the chord A G (Fig. 296) in a given case equal 20 feet, and the versed sine, b, or the rise DE, equal 4 feet; and let the ordinates be located at every 2 feet along the chord line, A G.

In solving this problem we require first to find the radius. This is obtained by means of equation (161.) -

a2+ b2/2b

For a, half the chord, we have 10 feet; for b, the versed sine, we have 4 feet; and, substituting these values, we have -

r =

102+42

116

14.5

2x4

8

The radius equals -

14.5

The versed sine equals -

4.0

(r-b)=

10.5

The square of 14.5, the radius, equals 210.25. Now we have, substituting these values in equation (163.) -

Section XI The Circle 410

The respective values of x, as above required, are 0, 2, 4, 6, 8 and 10 Substituting successively for x one of these values, we shall have, when -

Section XI The Circle 411

Values for t may be taken at points as numerous as desirable for accuracy.

In ordinary cases, however, they need not be nearer than in this example.

After the points are secured, let a flexible piece of wood be bent so as to coincide with at least four of the points at a time, and then draw the curve against the strip.

446. - Circle: Relation of Diameter to Circumference. - In Art. 439 it is shown that the area of a polygon equals the radius of the inscribed circle multiplied by half of a side of the polygon and by the number of the sides; or,