CHAPTER II

THE CIRCLE

14. Properties of the Circle-Simple Problems.--

(1) In a segment ABCD of a circle (Fig. 35) the angle AOD at the centre is double of the angle ABD at the circumference.

(2) Angles ABD and ACD in the same segment of a circle (Fig. 35) are equal to one another.

(3) The angle in a semicircle is a right angle. The carpenter, in cutting a semicircular groove in a piece of wood, makes use of this property when he tests it with a square, as shown in Fig. 36.

B

H

FIG. 35.

FIG. 36.

FIG. 37.

(4) A straight line which bisects a chord at right angles passes through the centre of the circle. This suggests a method of finding the centre of a circle to pass through three given points or to circumscribe a given triangle.

To describe an arc of a circle through three given points, A, B, and C, when the centre of the circle is inaccessible (Fig. 37). With centres A and C describe arcs CH3 and A3K. Join AB and CB, and produce these lines to meet the arcs at H and K. Mark off short equal arcs H1 and K1. The intersection P of the lines A1, C1, is another point on the arc required. In like manner other points may be found, and a fair curve drawn through all of them is the arc required.

The foregoing construction is based on the fact that the three angles of a triangle are together equal to 180°, and that the angles in the same segment of a circle are equal to one another; also that in equal circles equal arcs subtend equal angles at their centres. The student should have no difficulty in showing that the construction makes the angle APC equal to the angle ABC. For another method see Art. 59, p. 56.

(5) A tangent to a circle is at right angles to the radius or diameter drawn to the point of contact.

To draw a tangent to a circle from an external point. A good practical method is to place a straight-edge on the paper and adjust it so that the edge passes through the point and touches the circle, then the tangent may be drawn. The point of contact must, however, be obtained by dropping a perpendicular from the centre of the circle to the tangent.

The following construction is recommended when the one just given is objected to. With P (Fig. 38), the given external point as centre, describe the arc OSR, passing through O

the centre of the given circle. With centre O and radius equal to the diameter of the circle describe an arc to cut the arc OSR at R. Draw OR cutting the circle at T. PT is the tangent required and T is the point of contact. The student should satisfy himself that this construction makes the angle OTP a right angle.

S

FIG. 38.

Another method is to describe a semicircle on OP as diameter, to cut the given circle at T, which will be the point of contact of the tangent required. In this case the angle OTP is obviously a right angle because it is the angle in a semicircle.

It is obvious that two tangents may be drawn to a circle from an external point, and that they are equal in length.

To draw a tangent to two given circles. This may be done very accurately by placing a straight-edge on the paper, adjusting it so that the edge touches the circles, and then drawing the tangent. The points of contact must however be obtained by dropping perpendiculars from the centres of the circles to the tangent.

When the preceding construction is objected to, the following may be used: A and B (Figs. 39 and 40) are the centres of the given circles. Join AB cutting the circles at C and D. Make DE equal to BC. Draw

a circle with centre A and radius AE. Draw BF a tangent to this circle, F being the point of contact. Draw the line AF meeting the given circle, whose centre is A, at H. In Fig. 39, H is on AF produced. Draw BK parallel to AH, meeting the circle whose centre is B, at K.

C

The straight line joining H and K is a tangent to both the given circles, and H and K are the points of contact.

If the circles lie outside of one another four common tangents may be drawn to them.

(6) The angle ABC (Fig. 41) between the chord AB and the tangent BC to the circle at B is equal to the angle ADB in the alternate segment.

To inscribe in a circle ABD (Fig. 41) a triangle equiangular to the triangle EFH. Draw the tangent KBC. Draw the chord AB, making the angle ABC equal to the angle F. Draw the chord DB, making the angle DBK equal to the angle H. Join AD. The triangle ABD is the triangle required.

On a given line AB (Figs. 42 and 43) to describe a segment of a circle which shall contain an angle equal to a given angle C. Draw AD, making the angle DAB equal to the angle C. Draw AO at right angles to Bisect AB at E. Draw EO at right angles to AB, meeting AO at O. O is the centre, and OA is the radius of the circle of which the segment AFB contains an angle equal to the angle C.

AD.

(7) In a quadrilateral ABCD (Fig. 44) inscribed in a circle the opposite angles ABC and ADC are together equal to two right angles. Also the other pair of opposite angles are together equal to two right angles.

(8) In Fig. 44

AC × BD = AB × CD + AD× BC.

(9) AC and BD are any two chords of a circle

intersecting at F. AF CF = BF x DF. In

B

H

FIG. 44.

Fig. 44 the point F is within the circle, but it may be outside the circle.

(10) EH is a tangent to the circle ABCD (Fig. 44), H being the point of contact. EKL is a line cutting the circle at K and L.

EH2

= EK × EL. To draw a tangent to a given arc of a circle, through a given point, without using the centre of the circle.

Fig. 45.

The given point P is on the given arc APB but not near

either of its extremities. With centre P describe arcs of a circle to cut the given arc at A and B. A line through P parallel to the chord AB is the tangent required.

Let C be the middle point of the chord AB. If d is the diameter of the circle of which APB is an arc, then PC (d-PC)= ACX CB=BC2. BC2+PC2 PB2 PC

Hence d =

=

PC'

The lengths of PB and PC may be

E

measured and d found by arithmetic.

な

FIG. 45.

FIG. 46.

Fig. 46. The given point Q is on the given arc and near one of its extremities. Draw chords QP and QB. Join PB. Draw QE making the angle PQE equal to the angle PBQ. QE is the tangent required.

Fig. 47. The given point S is outside the given are. Draw the line SAB cutting the arc at A and

B.

12

=

Find a length such that SA. SB (see Art. 12, p. 11). With S as centre and radius equal to l, describe an arc of a circle to

FIG. 47.

cut the given arc at T. A line ST is the tangent required and T is the point of contact.

15. To construct a triangle having given the base, the vertical angle, and the length of the bisector of the vertical angle. First method (Fig. 48). On AB the given base describe a

segment of a circle ACB containing an angle equal to the given vertical angle. From O the centre of this circle draw OF at right angles to AB and produce it to meet the circle at D. On any straight line make ce equal to the given length of the bisector of the vertical angle. Bisect ce at g. Draw eh at right angles to ce and make eh equal the chord DB. With centre g and radius gh describe an arc of a circle to cut ce produced at d. With centre D and radius equal to

de describe an arc of a circle to cut the circle ACB at C. Join AC and BC. ACB is the triangle required.

The construction for finding the length of DC is based on the fact that DC X DE = DB2.

Second method (Fig. 49). Determine the circle ACB and the point D as in the first solution. Through D draw a straight line DQP cutting AB at Q. Make QP equal to the given length of the bisector of the vertical angle. Repeat this construction several times so as to determine a sufficient number of points on the locus of P. The intersection of the locus of P and the circle ACB determines the vertex of the required triangle.

This second method is a very good illustration of the use of a locus. It is quicker and probably more accurate and certainly much easier to discover.

16. To find the locus of a point which moves so that the ratio of its distances from two given points shall be equal to a given ratio.-Let A and B

(Fig. 50) be the two given points, and let P be one position of the moving point so that the ratio of AP to BP is equal to a given ratio. In Fig. 50 this ratio is 2:1.

Draw PD bisecting the angle APB and meeting AB at D. Draw PD, bisecting the angle between AP produced and BP. Bisect DD, at O. A circle with centre O and radius OD will be the locus required.

Draw BE parallel to PD to meet AP produced at E. parallel to PD, to meet AP at E.

It is easy to show that PE PE1

=

= PB.

Draw BE1

Also, AD: DB:: AP: PE, therefore AD: DB:: AP: PB, and D is a fixed point.

Again, AD, DB:: AP: PE,, therefore AD, D,B :: AP : PB, and D, is a fixed point.

The angle DPD, is obviously a right angle. described on DD, as diameter will pass through P.

Hence a circle

If the figure be rotated about AD, as an axis, the circle will describe the surface of a sphere, and the surface of this sphere will evidently be the locus of a point moving in space so that the ratio of its distances from the two fixed points A and B is equal to a given ratio.

17. The Inscribed and Escribed Circles of a Triangle. The inscribed circle of a triangle is the one which touches each of the three sides. An escribed circle touches one side and the other two sides produced. There are three escribed circles to a triangle. The constructions for drawing the inscribed and escribed circles are based on the fact, that when a circle touches each of two straight lines, its centre