A Key to the Exercises and Examples Contained in a Text-book of Euclid's Elements

II. ON THE TANGENT AND THE CONTACT OF CIRCLES.

Page 217.

1. Let AB be a chord of fixed length, P its middle point, and O the centre of the . Join OP. Then OP is perp. to AB [11. 3], and is of fixed length for all positions of AB [III. 14]; .. the locus of P is a concentric O, which is touched by AB at P, since AB is perp. to the radius OP [III. 16].

2. Let AP, AQ be two tangents drawn from A to a whose centre is O, and PR the diameter through P. Then shall the

PAQ be double of QPR. Join AO, cutting PQ at B. Then AO bisects the PAQ [III. 17, Cor.], and also bisects PQ at rt. angles [Ex. 2, p. 182]: also PR is perp. to AP [III. 18].

Hence from the rt. angled ▲ PAO, BPO, the

PAO = the BPO, each being the compt. of the POA. .. the PAQ is double of the

QPR.

3. Let A be the point of contact of the two O, PAQ the st. line through A terminated by the ces, and PB, QC the tangents at P, Q. Then shall PB, QC be par1. Through A draw BAC perp. to the line of centres, meeting PB, QC at B and C. Then BAC at A [III. 16]. And because BP: .. the BPA = the BAP = the vert. opp. (since CA = CQ). That is, the BPQ = the

touches both

par1. [1. 27].

BA [III. 7, Cor.], CAQ = the CQA

CQP; .. PB, QC are

4. Let A be a point of intersection of the two O, PAQ the st. line through A terminated by the Oces, and PR, QR the tangents at P, Q: and let the tangents at A meet PR, QR at B, C.

Then shall the PRQ = the BAC. For, from the two isosceles ABPA, CAQ [III. 17, Cor.], the BPA = the BAP, and the CQA = the CAQ: .. the two 4 RPQ, RQP together the two ▲ BAP, CAQ. Hence the PRQ= the BAC [I. 32, 1. 13].

5. Let the two par1. tangents AP, BQ touch the C at A, B, and cut off the segment PQ from a third tangent whose point of contact is R. Take C the centre. Then shall the PCQ be a rt. L. Join CA, CB, CR.

line.

Then CA, CB being perp. [III. 18] to par1. lines are in one st. And PC, QC respectively bisect the

Cor.]. Hence the

PCQ is half of the four

of two rt. angles [1. 13].

H. K. E.

ACR, BCR [III. 17, at C, that is, half

6. Let O be the centre of the circle; B, C the points of contact of the fixed tangents AP, AQ, and R the point of contact of the third tangent PQ. Then shall the POQ be constant. Join OB, OR, OC. Now OP, OQ respectively bisect the ▲ BOR, COR [III. 17, Cor.]; .. the POQ is half the reflex which is constant, for B, O, C are fixed points.

NOTE. The Ex. 36, p. 228.]

BOC,

POQ = one rt. angle + half the at A. [See

7. Let ABCD be the quad'., and P, Q, R, S the points of contact of the sides AB, BC, CD, DA.

Then ASAP [III. 17, Cor.] and DS=DR; .. by addition AD=AP, DR. Similarly BC= BP, CR. Hence AD and BC together =AP, BP, DR, CR = AB, DC.

8. Let ABCD be a quad'. in which AB, CD together BC, DA. By bisecting the two ABC, BCD describe a to touch three Then shall AD also touch this

sides AB, BC, CD [Ex. 1, p. 182].

Also by Ex. 7,

. For if not, from A draw AD' touching the O and cutting CD at D'. Now by hyp., AB, CD together BC, AD. AB, CD' together = BC, AD'. BC, AD'. .. taking the differences of these equals, DD' the difference of AD and AD'; hence either AD' = AD, DD', or AD = AD', DD'; which is impossible [1. 20].

9. Let A be the point of contact, B the centre of the inner C, C of the outer O. Then A, B, C are collinear [III. 11]. Let BC, produced if necessary, cut the inner ce at D. Let EF be the chord of the outer which touches the inner at D, and is therefore perp. to AD [III. 18]; and let PQ be any other chord touching the inner O. From C draw CR perp. to PQ: then R is outside the inner [III. Def. 10]. Let CR cut the Oce at S. Now CS is greater than CD [III. 7]; much more is CR greater than CD; .. EF is greater than PQ [III. 15].

10. Let ABC be a ▲, and F the middle point of the side AB. On BC as diameter describe a ○ ; call its centre D. Join FD and produce it to meet the ce at P. Then FP is made up of FD and DP; of which FD is half AC [Ex. 3, p. 97] and DP is half BC: that is FP is half the sum of BC, CA. And a described from centre F with radius FP will touch the C on BC at P; for the centres of the two circles and the point P are collinear. Similarly, the same will touch the on AC.

11. Let A be the given point, O the centre of the given O, and X the given st. line. In the place a chord PQ equal to X see IV. 1]. With centre O, and radius equal to the perp. from O on PQ, describe a circle, which will be touched by PQ [III. 16]. From A draw ABC to touch the inner [III. 17] and to cut the given at B, C. Then BC = PQ, being chords at equal distances from the centre of the given [III. 18 and 14].

If A is without the O, X must be not greater than the diameter. If A is within the O, X must be not greater than the diameter, and not less than the chord drawn through A perp. to

OA.

Draw AP

12. Let O be the given point in the given st. line; and let AB, the given par'., cut any of the system at A, B. the tangent at A, and OP perp. to AP. Take C the centre of the O, and join AC, OC: then OC cuts AB in R at rt. angles [Hyp. and I. 29].

Then the L POA =

the ROA, for each is equal to the OAC [1. 29, 1. 5]. Hence ▲ AOP, AOR are identically equal by 1. 26. So that OP=OR; and OR is constant, for all of the system. Now AP is perp. to OP... AP touches the fixed is O and radius OR.

whose centre

13. Let A be the centre of the outer, and B of the inner fixed O. Let P be the centre of any third touching the first C at D and the second at E. Then shall AP + BP be constant. Let T1, r2, r, denote the radii of the three Then the points A, P, D and B, E, P are collinear [111. 11 and 12]. And

AP+BP = 1 − r 3 + r2 + r3 = r1 + r2•

NOTE. This problem is a special case of the following: If any two circles are touched one internally and one externally by a third circle, the sum or difference of the distances of this third circle from the centres of the given circles is constant.

Now CP

14. Let PA, PB be any pair of tangents containing the given angle. Take C the centre of the O, and join CA, CP. bisects the APB [III. 17, Cor.].

Hence in the ACAP, the - CAP, CPA, and the side CA are constant; .. CP is constant [1. 26]. .. the locus of P is a concentric .

се

15. Let A and B be the centres of the two given ©o, X and Y the given st. lines. At any point C on the Oce of the first draw a tangent CP equal to X. From A as centre with radius

AP describe a O, and shew that its Oce is the locus of points from which tangents of the required length may be drawn to first given circle. Proceeding in a similar manner with the second circle, we see that the points common to the two locus-circles satisfy the conditions. There are two solutions, one solution or no solution according as the loci-circles intersect, touch one another, or do not meet.

16. Lemma. If ABC is a triangle, and X a point in the base, such that AB2~ AC2 BX2 ~ CX2, then AX is perp, to BC. This is the converse of Ex. 7, p. 84, and may be proved indirectly from that theorem.

Then BC, CA,

Let A, B, C be the centres of the three ©3. AB pass respectively through P, Q, R the points of contact [III. 12]. Let the common tangents at Q and R meet at O. Join OP. Then OP shall touch the 3 (B) and (C) at P.

OB, OC.

Join OA,

Now OB2 = OR2 + BR2, and OC2 = OQ2 + CQ2 [111. 18, 1. 47] Hence by subtraction, remembering that OQ=OR [III. 17, Cor.}

OB2 ~ OC2 = BR2 ~ CQ2 = BP2 ~ CP2.

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17. (i) The two direct tangents only can be drawn in this case for when we attempt to draw the transverse tangents we find the point B within the circle of construction.

.. no tangent can be drawn to it from B.

(ii) Here the two direct tangents may be drawn, and the two transverse tangents become coincident. For B will fall on thece of the circle of construction; hence only one tangent (or two coincident tangents) may be drawn to it from B.

(iii) Hence for similar reasons the two direct tangents are coincident and the two transverse tangents are impossible.

(iv) Both direct and transverse tangents are impossible.

18. In this case the of constr. is reduced to a point. Proceed thus:-join AB the centres of the given C, and draw AD, BE perp. to AB, cutting the Oces in D and E. Join DE, which will be one direct common tangent. [Proof by 1. 28, 33, 34, and III. 16.]

19. Let a pair of common tangents touch the greater at D, D', the smaller at E, E', and cut one another at P.

Then by III. 17, Cor., PD = PD', and PE = PE'. .. for direct tangents PD - PE = PD' – PE'; and for transverse tangents PD + PE = PD' + PE'; .. in either case

DE = D'E'.

If the are equal, then the direct common tangents are equal [1. 34]. Or again, with the fig. of p. 218, DE larly D'E' = BC'; but BC = BC', .'. DE = D'E'.

BC; simi

20. Let the direct common tangents DE, D'E' touch the whose centres are A, B at D, E and D', E', and cut one another at P. Join PB, BE, BE'. Then in the ▲ PEB, PE'B, we have BE = BE' and BP common, also the PEB, PE'B are rt. 4o [III. 18];

.. ▲ EPB = ▲ E'PB [Ex. 12, p. 91].

That is, the centre B lies on the bisector of the

between the

common tangents. Similarly the centre A lies on the same bisector. Therefore the points A, B, P are collinear.

21. Let B, C be the centres of the two given : then BC passes through A [III. 12]. Join BP, CQ.