chord the space included between those lines and the arc is termed a sector. The diameter A O B divides the circle into two equal parts, each of which is called a semicircle. The perpendicular OF upon AB divides the semicircle into two equal parts AF, BF, called quadrants. Circles are said to touch each other when their circumferences meet at a point but do not cut one another, and the right line joining their centres passes through the point of contact. Two or more circles are said to be inscribed in another circle when they touch its circumference on the inner or concave side, and also touch each other on the outer or convex side. Two circles drawn from the same centre but with different radii are said to be concentric.

A rectilinear figure is inscribed in a circle when all its angular points, or vertices, are on the circumference of the circle, and the circle is then said to be circumscribed about the rectilinear figure. A rectilinear figure is circumscribed about a circle when all its sides are tangents thereto, and the circle is then said to be inscribed in the figure.

48. To draw a tangent at a given point of a given circle.Let O (fig. 40) be the centre of the circle, A the given point on its circumference; join A O. At the point A draw AT perpendicular to AO; then AT is the tangent required.

49. To describe a circle of given radius touching a given circle in a given point.-Let O (fig. 41) be the centre of the given circle, A the point on its circumference where the contact is to take place. Draw the radius OA and produce it to Q, making AQ equal to the given radius. From Q as a centre, with QA as radius, describe a circle which will touch the given one at the point A as required.

If the given radius is less than that of the given circle,

cut off from A O the length AP equal to the given radius, and from P as a centre, with PA as radius,

describe a circle whose convex side will touch the given one on the concave side of its circumference at the point A.

If P is the centre of the given circle, and the given radius is greater than PA, produce A P to O, making A O equal to the given radius. Then a circle drawn from O as a centre, with OA as radius, will touch the given one on the convex side of its circumference.

If AT is drawn perpendicular to OAQ it will be the common tangent of the two circles which are in contact at A.

50. To describe a circle through three given points which are not in the same right line.-Let D, C and F (fig. 41) be the three given points. Join CD and CF, and bisect CD and CF at E and B. Draw EO at right angles to CD, BO at right angles to CF; these will meet at the point O which is the centre of the required circle. Join OC; then a circle described from O as a centre, with OC as radius, will pass through the three given points.

51. From a given point outside a given circle to draw

tangents thereto.-Let O (fig. 42) be the centre of the given

Fig. 42.

circle, O A its radius, T the given

point outside the circle. Join
OT; then from O as a centre,
with OT as a radius, describe
another circle concentric with the
given one.
Let B be the point
where OT cuts the given circle,
draw DBC at right angles to

OB, and cutting the outer circle
in D and C. Draw the lines OD,

OC, cutting the given circle in E and A; join TE, TA; then TE and T A are the required tangents at the points E and A of the given circle.

Another method.

Fig. 43.

Let O (fig. 43) be the centre of the given circle, T the given point. Draw the diameter HOA and produce it to T and B; making TB equal to T A. Bisect HB in C, and from C as a centre with CB as radius, describe a semicircle. Draw TD at right angles to HB, meeting the semicircle in D: from T as a centre, and with TD for radius, draw an arc, cut

ting the given circle in E and F. Join OE, OF, TE, TF; then TE and TF are the tangents required.

This solution depends on the property that the rectangle under HT and TA is equal to the square on the tangent TE or TF, and also that the rectangle under HT and TB is equal to the square on TD (36)

52. To find any number of points upon an arc of a circle, the middle point and two extremities of the arc being given.-Let A and B (fig. 44) be the two extremities of the

arc, D the middle point. Join A B; and bisect the chord AB in C; draw DC which will be perpendicular to A B. Draw the chords A D, D B, and divide DB into any number of parts by the points E, F, &c. From B as a centre, and with BA as radius, describe a circle AGH. Draw EG, FH parallel to DA, cutting the circle in G and H. From B as a centre, and with radius BE, describe a circle; and from A as a centre, with GE as radius, draw an arc cutting the last circle in P. Then P is a point on the required arc. In the same manner from B as a centre, and with radius BF, describe a circle; and from A as a centre, with HF as radius, draw an arc cutting this circle in Q; then Q is also a point on the required arc. In this way any number of points can be found upon the arc of a circle when the centre is at too great a distance to be available; the arc itself can either be drawn by hand or by help of a bent lath passing through all the points.

This method is derived from the principle that all the

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angles in the same segment of a circle are equal; for the angles A D B, A PB, A QB, are all equal. The entasis or curved outline of the shaft of a column or the contour of a spire can be drawn on a large scale in this way.

53. To draw an arc of a circle by continuous motion, when the centre is not available.-Take two straight rulers of twice the length of the chords A D, BD (fig. 44), and fasten them together so as to make with each other an angle equal to A D B, their outer edges being applied to the chords A D, DB. Fix a pin at A and another at B against the outer edges of the rulers; let a pencil be fixed at the vertex D, and as the rulers move backwards or forwards against the pins the pencil at D will mark out the arc of the circle. Another method.

Let BC (fig. 45) be the chord of the arc, A G its height Fig. 45.

at the centre, EAD the tangent at A perpendicular to A G, and parallel to BC. Draw the chord A B, and take A D equal to AB; join BD. Now let the triangle BAD be made rigid, and let it move about pins fixed at A and B ; then a pencil fixed at its vertex A will mark out the arc AFB as the triangle moves from right to left. So also by making the triangle move from left to right we can trace the arc A C.

This method is based on the principle that if we take any point F in the arc A B, and draw the chords A F, B F, then the angle BFA equals the angle B A D.