lies on the line bisecting the angle between them. In Fig. 51, O is the centre of the inscribed circle of the triangle ABC, and O, is the centre of one of the escribed circles.

The following results are not difficult to prove :-

(1) The sides of the triangle formed by joining the centres of the escribed circles pass through the angular points of the original triangle.

B

FIG. 51.

(2) The line joining the centre of one of the escribed circles and the opposite angle of the triangle passes through the centre of the inscribed circle and is perpendicular to the line joining the centres of the other two escribed circles.

=

=

(3) Referring to Fig. 51, AE AF half the perimeter of the triangle ABC. This suggests the construction for the solution of the following problem.

Given the perimeter1 and one angle of a triangle, also the radius of the inscribed circle to construct the triangle. Make the angle EAF (Fig. 51) equal to the given angle. Draw a circle having a radius equal to the given radius to touch AE and AF. This will be the inscribed circle of the required triangle. Make AE and AF each equal to half the given perimeter. From E and F draw perpendiculars to AE and AF respectively to meet at O1. A circle with O, as centre and OE or O,F as radius will be an escribed circle of the triangle. A tangent BC to the two circles now drawn will complete the triangle required.

18. Circles in Contact.-In considering problems on circles in contact with one another two simple facts should be kept in view, viz. (1) When two circles touch one another the straight line which joins their centres or that line produced passes through the point of contact.

(2) When two circles touch one another the distance between their centres is equal to either the sum or difference of their radii.

To draw a circle of given radius to touch two given circles. Three cases are shown in Figs. 52, 53, and 54. A and B are the centres of the given

K

K

FIG. 52.

FIG. 53.

circles. AB or AB produced cuts the given

CD and FE each equal to the given radius.

FIG. 54.

circles at C and F. Make With centre A and radius

AD draw the arc DO. With centre B and radius BE draw the arc

The perimeter of a triangle is the sum of its sides.

EO, cutting the former arc at O. Join O with A and B, and produce these lines if necessary to meet the given circles at H and K. O is the centre of the required circle, and H and K are the points of contact. 19. To draw a series of circles to touch one another and two given lines.-AB and CD (Fig. 55) are the two given lines. Draw EF, bisecting the angle between AB and CD. Let E be the centre of one circle: its radius is EA, the perpendicular on AB from E. Draw HK perpendicular to EF. Make KL equal to KH. Draw LM perpendicular to AB to meet EF at M. M is the centre and ML is the radius of the next circle.

A

FIG. 55.

FIG. 56.

E

20. To draw a circle to pass through a given point and touch two given lines.-AB and AC (Fig. 56) are the two given lines and D is the given point. Draw AE bisecting the angle BAC. Join AD. Take any point F in AE. Draw FH perpendicular to AB. With F as centre and FH as radius describe a circle. This circle will touch the lines AB and AC. Let this circle cut AD at K. Draw DO parallel to KF, meeting AE at O. O is the centre and OD the radius of the circle required. 21. To draw a circle to touch two given lines and a given circle.-AB and CD (Fig. 57) are the given lines and EF is the given circle, N being its centre. Draw HK and LM parallel to AB and CD respectively, and at distances from them equal to the radius EN of the given circle. Draw, by preceding problem, a circle to pass through N, and touch the lines HK and LM. O the centre of this circle is the centre of the circle required, and OE is its radius, OEN being a straight line.

H

K

A

B

EN

D

M

FIG. 57.

line and pass

22. To draw a circle to touch a given through two given points.-AB (Fig. 58) is the given line, and C and D are the given points. Draw CD and produce it to meet AB at E. If the required circle touches AB at K, then EK2 = ED × EC, or EK must be a mean proportional to EC and ED. Hence the following construction. Produce CE and make EF equal to ED. On CF describe a semicircle. Draw EH perpendicular to CF to meet the semicircle at H. Make EK equal to EH. Draw KO perpendicular to AB, and draw LO, bisecting CD at right angles to meet KO at O. O is the centre of the circle required.

FIG. 58.

B

D

23. To draw a circle to touch a given circle and a given line at a given point in it.-ABC (Fig. 59) is the given circle, DE is the given line, and D the given point in it. Through F, the centre of the given circle, draw FE perpendicular to DE and produce it to meet the circle at C. Draw DO perpendicular to DE. Draw CD cutting the circle at B. Draw FB and produce it to meet DO at O. O is the centre and OD the radius of the circle required. There are two solutions, the second being obtained in the same way by joining A with D instead of joining C with D.

E

FIG. 59.

24. To draw a circle to pass through two given points and touch a given circle.-A and B (Fig. 60) are the given points and CDE is the given circle. Draw a circle CABE through A and B, cutting the circle CDE at C and E. Join CE, and produce it to meet AB produced at F. Draw FD, touching the given circle at D. Join H, the centre of the given circle, with D, and produce it to meet a line bisecting AB at right angles at O. O is the centre of the required circle. There are two solutions. The second is obtained by drawing the other tangent to the given circle from F and proceeding as before.

FIG. 60.

Considering a straight line to be a circle of infinite diameter, the student should endeavour to deduce from the foregoing solution the construction for drawing a circle to pass through two given points and touch a given line.

25. To draw the locus of the centre of a circle which touches two given circles.-A and B (Fig. 61) are the centres of the given circles. Join AB cutting the cir

cles at C and D.

Bisect CD at E. Mark

off from E, above and below it, on AB a number of equal divisions. With centres A and B describe arcs of circles through these divisions as shown. A fair curve drawn through the intersections of these arcs as shown is the locus required. The curve is an hyperbola whose foci are the points A and B.

FIG. 61.

This locus may be used in solving problems on the drawing of a circle to touch two given circles and to fulfil some other condition.

When the given circles are external to one another, as in Fig. 61, four different loci may be drawn in the manner described above, because there are four positions of the point E. If AB produced cuts the upper circle again in F and the lower one again in H, then the remaining positions of the point E are, the middle point of FH, the middle point of FD, and the middle point of HC.

A

B

26. To draw the locus of the centre of a circle which touches a given line and a given circle.-AB (Fig. 62) is the given line and CDE is the given circle, O being its centre. Through O draw OH perpendicular to AB, meeting AB at H, and the given circle at D. Bisect DH at F. Mark off from F above and below it, on OH, a number of equal divisions. With centre O draw arcs through the divisions above F to meet parallels to AB through the corresponding divisions below F, as shown. A fair curve drawn through the intersections of the arcs and parallels, as shown, is the locus required. The curve is a parabola.

FIG. 62.

This locus may be used in solving problems on the drawing of a circle to touch a given circle and a given straight line, and to fulfil some other condition.

Two different loci may be drawn in the manner described above, because there are two positions of the point F. If the line OH produced cuts the circle again at K, then the second position of F is the middle point of HK.

27. Pole and Polar.-If through a given point P (Figs. 63 and 64) any straight line be drawn to cut a circle NQR at Q and R, tangents to the circle at Q and R will intersect on a fixed straight line LM. Conversely, the chord of contact of tangents from any point in LM, outside the circle, will pass through P. This fixed line LM is called the polar of the point P with respect to the circle, and the point P is called the pole of the line LM with respect to the circle.

LM will be found to be perpendicular to OP, O being the centre of the circle, and if OP or OP produced cuts LM at T, and the circle at S, then OP × OT

=

OS2.

When the pole P is outside the circle (Fig. 64) the polar is the chord of contact of the tangents from P to the circle.

If chords QR and NK of a circle meet (produced if necessary) at P, then the chords NQ and KR will intersect on the polar of P, also the chords NR and QK will intersect on the polar of P. This suggests a construction for drawing the polar of a point P by using a pencil and

straight-edge only; and, having found the polar, lines from P to where the polar cuts the circle will be the tangents to the circle from P. Tangents to a circle from an external point may therefore be drawn, and their points of contact determined, by using a pencil and straightedge only.

28. Centres of Similitude.-CD and EF (Figs. 65 and 66) are two parallel diameters of two circles whose centres are A and B. The

straight lines joining the extremities of these diameters intersect in pairs at fixed points S and S, on the line joining the centres of the circles. The points S and S, are called the centres of similitude of the two circles, and SA : SB :: AC: BE, also S,A: S,B :: AC: BE.

When the circles are external to one another, as in Fig. 66, one pair of the common tangents will intersect at S and the other pair will intersect at S1.

When the two circles touch one another one of the centres of similitude will be at the point of contact of the circles.

29. The Radical Axis.-The locus of a point from which equal tangents can be drawn to two given circles is a straight line called the radical axis of the two circles.

If the circles touch one another the common tangent at their point of contact is their radical axis, and if the circles cut one another their common chord produced is their radical axis.

For any other case the following general construction may be used for finding the position of the radical axis of two circles. A and B (Figs. 67 and 68) are the centres of the given circles ECF and HDK. Describe any circle to cut the former circle at E and F and the latter at H and K. Draw the chords EF and HK and produce them to meet at P. A line PO at right angles to AB is the radical axis required. The proof is as follows: Since E, F, H, and K are on the same circle, PE × PF = PH × PK. But PC being a tangent to the circle ECF, PC2 = PE x PF. Also PD being a tangent to the circle HDK, PD2 = PH x PK; therefore PC2 = PD2 and PC = PD. Hence P is a point on the radical axis.

It can be proved that AO2- BO2 = AC2 BD2.

If a circle be described with its centre at any point P on the radical