Suppose that the arm to the wheel O' (fig. 189), instead of being fixed by a button at A', is attached

by a button and groove at B. It is evident that the point B will move backwards and forwards on the arc

bc. Conversely, an oscillating movement on bc, will produce the rotation of OA. Figure 192 shows how the rectilinear movement of the shaft T produces the rotation of the wheels, O and O'.

Questions for Examination.

1. Define a secant.

2. Prove that the perpendicular from the centre on a chord is less than the radius.

3. Prove that the perpendicular from the centre on a chord bisects the chord and the arc it subtends.

4. Prove that the circle is the only curve in which every diameter is a symmetrical axis.

5. Describe the curved spirit-level, and show how it may used to find the angle of inclination of a surface.

6. Describe a circle,

be

Ist. From a given centre, passing through a given point. 2nd. Of given radius passing through two given points. When is this case impossible?

3rd. Through three given points.

4th. Through the angular points of a triangle.

7. Find the centre of a given circle.

8. To bisect a given arc.

9. To bisect an angle.

10. Having given the angle in a segment of a circle, and the chord which subtends it, describe the arc-first, when the space on the other side of the chord is unlimited; and, second, when it is limited.

II. Prove that parallel secants intercept equal arcs.

12. Show how to fit, together the beading of a cupboard, the angle of which is given.

13. An inscribed angle is half the angle at the centre on the

same arc.

14. Every angle described in a semicircle is a right-angle.

15. Every segment of a circle, the inscribed angle of which is a right-angle, is a semicircle. Mention any application of this property you may have observed.

16. If the chords A B, A'B' intersect in a point O, then— OA:O A'=O B': OB,

whether O be within or without the circle.

17. Prove that half a chord perpendicular to a diameter is a mean proportional between the segments of the diameter.

18. In two circles the extremities of two parallel radii are joined; prove that the ratio of the parts into which they divide the line joining the centres is constant.

Theorems and Problems.

1. Describe a circle passing through the angular points of a square.

2. Describe a circle passing through the angular points of an isosceles triangle.

3. Having given an arc of a circle, complete it.

4. Construct angles of the following dimensions ; namely, 90°, 45°, 60°, 75°, 30°, 15°, 135°, 120° and 105°.

5. The locus of the apex of every triangle on the same base and having the same vertical angle is an arc of a circle.

6. If any quadrilateral have its angular points on the circumference of a circle, its opposite angles are together equal to 180.°

7. If a chord be not a diameter it divides the circle into two unequal segments, and an angle inscribed in the one greater than the semicircle is acute, while the angle inscribed in the other is obtuse.

8. Angles at the circumference on the same arc are equal.

9. Construct a triangle, having given the base, difference of the sides, and one of the angles at the base.

10. Construct a triangle, having given the base the difference of the sides, and difference of the angles at the base.

II. Construct an isosceles triangle, having given the vertical angle and a point in the base.

12. Construct a triangle, having given

Ist. Two sides, and a line joining the middle point of base with the opposite angle.

2nd. Two sides, and a line joining the middle point of one of them with the opposite angle.

3rd. The lines drawn from the middle points of the sides to the opposite angles.

4th. The feet of the three heights.

5th. One angle, the perimeter, and the height from the angle.

6th. One angle, the perimeter, and the height opposite the angle.

7th. A side, an adjacent angle, and the length of the bisector of this angle.

8th. The sum of two sides and the angles.

9th. The perimeter and the angles.

10th. An angle, the length of the bisector, and the height from this angle.

11th. Ditto, the height being opposite the angle.

12th. The base, the sum of the two sides, and the difference of the angles at the base.

13th. The perimeter, the size and position of one angle, and a point in the opposite side.

14th. A side, an angle, and a height (five cases).

13. Construct a right-angled triangle, having given

Ist. One of the sides and the excess of the hypothenuse over the other side.

2nd. The angles and the difference of the hypothenuse and one of the other sides.

Arithmetical Questions.

I. An angle at the centre of a circle is 38°; find the angle at the circumference on the same arc.

2. An angle at the circumference is 54°; find the angle at the centre on the same arc.

3. An angle at the circumference is 34°; find the degrees in the arc intercepted by diameters parallel to its sides.

4. The side of a square formed by joining the extremities of two perpendicular diameters measures 7 ft. 6 in., find the length of the diameter of the circle. Ans. 10 60659 ft.

5 Through a point O within a circle, two straight lines A B, A B' are drawn to the circumference; A B-18 in., A O=12, A'O-3 in. Find A' B'. Ans. 24 in.

6. If in the above case A B=31 in., A' B'=32 in., and O B= 15, find the length of the segments of A'B'.

Ans. 20 and 12 in.

CHAPTER IX.

TANGENTS.

216. A tangent to a circle is a straight line which has one point in common with the circumference, and only The tangent to a circle has many important properties. Thus :

one.

Every straight line perpendicular to the extremity of a diameter is a tangent to the circumference. 217. Let A T be a perpendicular to the extremity A

A M

Fig. 193.

of the diameter (fig. 193); join with the centre any point M of this perpendicular; the line O M is an oblique line with respect to the straight line A T, wherefore it is longer than the perpendicular O A, which is the radius of the circle. Therefore M is outside

the circumference.

Hence the point A is the only point of the line AT in the circumference; hence it follows from the definition that this straight line is a tangent.

The point A is called the point of contact. The following problems naturally arise in this place.

To draw a tangent.

1st. From a given point in the circumference.

218. It is sufficient to erect a perpendicular at the extremity of the radius.

2nd. Through a given point without the circle.

Join the given point A to the centre O (fig. 194), and upon this line O A, as diameter, describe a circle. It cuts the given circle in two points,

D and D', which will be the points of contact of the tangents required; hence, by joining the points AD, A D', we shall have two tangents through the point A, for the angles ADO, A D'O, inscribed in a semicircumference, are right-angles.

We might, however, draw the tangent without sensible error by applying a ruler to the point A, and moving it so that it touches the circle. This method may be as exact practically as the first, and

Fig.194.

even more so, for the first construction, although theoretically correct, requires various operations, the accuracy of which depends upon the skill of the operator and the perfection of his instruments.

Thus error may be made in determining the middle point of A O, for example, for there are three operations to find this point, and each is liable to error; precision may also be wanting in describing the circle upon A O as diameter.

The figure shows that the diameter A O is a symmetrical axis. Therefore,