7. If possible, let a st. line cut the in the three points A, E, B, whereof E is between A and B. Then E must fall within the O. But it was assumed to be on the circumference. Hence the st. line AB cannot cut the O in more than two points. 8. Draw a chord perp. to the st. line joining the given pt. to the centre. This chord will be bisected at the given pt. 9. Let O be the common centre, ABCD the st. line cutting the inner in B, C, and the outer in A, D. Draw OX perp. to ABCD. Then BXCX and AX = DX. ference CD. .. difference AB = dif 10. The st. line, through the centre, perp. to one of the par1. chords, is perp. to the other [1. 29]. And this st. line bisects both chords. Hence, the st. line joining the middle pts. of two par1. chords passes through the centre. 11. The st. line, through the centre, perp. to one of the par1. chords, is perp. to all of them [1. 29]. And this st. line bisects all the chords. Hence it is the required locus. 12. Let the two Os intersect in A, B. Let CAD, EBF be par1. st. lines cutting the one circle in C, E and the other in D, F. Then the st. line through the centre of ABEC perp. to the chords AC, BE bisects these chords in P and Q, say. Similarly the st. line through the centre of ABFD perp. to the chords AD, BF bisects these chords in X and Y, say. But PQ is par1. to XY [1. 28]. .. PX = QY [1. 34]. But CA is double of PA, and AD is double of AX. .. CD is double of PX. Similarly EF is double of QY, .. CD = EF. 0. 13. Bisect PQ in R, and XY in Z. Let PY, QX intersect in Join OR, OZ. Then APXY = ▲ QXY; and AXOZ = AYOZ. .. ΔΡΧΟ = ΔQYO. Also ΔΡOR = ΔQOR. :. Δ POR, ΡΧΟ, ΧΟΖ together =▲ QOR, QYO, YOZ: i.e. the lines OR, OZ bisect the trapezium PXYQ. But the st. line RZ bisects the trapezium PXYQ [Ex. 8, p. 109, 1. 38]. .. the st. line RZ coincides with the st. lines OR, OZ: that is, o lies on RZ. Similarly if PX, QY intersect in O', o' lies on RZ. Page 158. 1. The diagonals of a parm. bisect one another. .. their pt. of intersection is the centre. Then the [Ex. 1]. And because AO = DO, 2. Let the parm. ABCD be inscribed in a . diagonals AC, BD intersect in O, the centre of the Because AO BO, .'. ▲ OAB = L OBA. = .. LOAD= L ODA; .. whole DAB = sum of $ ABD, ADB; = 3. Let C, D be the centres of two O intersecting in A. Draw AX perp. to CD, and produce it to B, so that BX AX. Then CA CB and DA = DB [1. 4]. .. B is a pt. on both circles. Page 168. 1. Let A be the centre of the larger, B of the smaller . Produce AB to C, making BC= the radius of the smaller . Then AC is the radius of the larger. .. the meet at C. Let D be any other pt. on the smaller . Then BD = BC... AB, BD together = AC. But AB, BD together > AD. .. ACAD. ... D cannot be on the circle with centre A and radius AC. = BQ, 2. Let C be the pt. of contact. Then ABC is a st. line [III. 11]. Because AC = AP, therefore ▲ ACP = ▲ APC. And because BC .. ▲ BCQ = ▲ BQC. .. ▲ APC = BQC... AP, BQ are parallel. Page 169. 1. The required locus is the st. line joining the centre of the given to the given pt. [III. 11, 12]. 2. On Let O be the centre of the given O, OP its radius. OP take PC equal to the given radius of the circles which are to touch the given O. Then the with centre C and radius CP will touch the given at P. And OC = the sum or the difference of OP and CP. Hence the required locus is a circle with centre O and radius equal to the sum or the difference of the radius of the given circle and the given radius of the touching circles. with with 3. Let A, B be the centres of the two circles. From AB cut off AC = radius of with centre A. Then BC= radius of centre B... the Os meet at C. Let D be any other pt. on centre B. Then AD, DB together > AB. But BD = BC. .. AD > AC. .. D cannot be on the circle with centre A and radius AC. 4. Let C be the pt. 12]. Because AC = AP, .. ▲ ACP = 4 APC. .. 4 BCQ = L BQC... APC = L BQC. of contact. Then ACB is a st. line [III. And because BC = BQ, .'. AP, BQ are parallel. Page 171. 1. Let A, B be the two given pts. Bisect AB in C : draw CX perp. to AB. Then, if CX coincides with the given st. line, with any pt. X on CX as centre, the circle described with centre C and radius CA will pass through B. But if CX is par1. to the given st. line no circle can be described as required. Finally, if CX cuts the given st. line in X, the circle described with centre X and radius CA or CB is the circle required. = Draw AM 2. Let A be the given pt., XY the given st. line. perp. to XY, and produce it to B, so that BM AM. All the circles will pass through B. [See Ex. 1, [See Ex. 1, p. 215.] 3. Let A be the centre of the given O, P the given pt. Take PC equal to the given radius, either on PA (produced if necessary) or on AP produced. The circles described with centre C and radius CP will touch the given circle at P: but, if the given radius is equal to the radius of the given circle, one of the two circles so described will coincide with the given circle. 4. Let A be the centre of the given O, B the given pt. Let AB cut the given C in C and D. The described with centre B and radius BC or BD will touch the given . Hence there are two solutions except when B is on the ce of the given O. 5. Let A be the centre of the given O, B the given pt. on it, C the given pt. through which the required is to pass. Let the perp. bisector of BC cut AB in O. The ✪ described with centre O and radius OB is the required: but, if C is on the circumference of the given O, O will coincide with A [III. 1], and the described will coincide with the given O. The solution is also impossible, if CB is perp. to AB. For then the perp. bisector of BC will not cut AB. S 6. Let A and B be the centres of the two given X and Y. Describe a circle with centre A and radius equal to the sum or the difference of the radius of the required circle and the radius of X. Describe a circle with centre B and radius equal to the sum or the difference of the radius of the required circle and the radius of Y. Then the circle described with either of the pts. of intersection of these two circles as centre and with the required radius will be the circle required. There are thus in general 8 possible solutions. 7. Let A, B be the middle pts. of the given chords DAF, EBG, so that AD 3 in. and BE = 4 in. Then, supposing the chords to be on the same side of the centre, let AB be produced to C the centre. Then CD2=CE2; that is, CA2 + 9 = CB2 + 16. .. CA - CB2 = 7. But CA - CB = 1. .. CA + CB, that is, AB + 2CB=7. ... CB = 3. .. CE2 = CB2 + BE2 = 32 + 42 = 25. .. CE (the radius) = 5. 8. Let A, B be the centres of the two circles, touching externally at C. Then ACB is a st. line [111. 12]. Draw the parallel diameters DAE, FBG. Then, because AD = AC, .. LADC= 2 ACD, .. ext. EAC = twice ACD. Similarly, CBF = twice BCG. But EAC = CBF, because AE, BF are par1... ACD = 1 BCG. .. DC, CG are in a st. line; that is, the st. line joining GD passes through C. Then 9. Let A, B be the centres of the two Os; and D, E the pts. where AB cuts the circles. Let PQ be any other st. line cutting the circles in P and Q. Let QA cut the 'A' circle in R. QP > QR (III. 8). Also AQ > AE, of which AR = AD; .'. QR> DE. .. à fortiori, QP > DE. .. DE is less than any st. line PQ cutting the in P and Q. Similarly, if AB produced cut the circles in FG, FG is greater than any of the st. lines PQ cutting the 3 in P and Q. S 10. Let BC be a chord: A the centre, and ADE the radius bisecting BC at rt. s in D. Let G be any pt. in BD, and GF perp. to BD, cutting the arc BEC in F. Draw AH perp. to FG. Then AE AF > HF. And AD = HG ; .. the whole or remainder DE > whole or remainder GF. In DG take a pt. L, and draw MLK perp. to BC to meet the arc at K. Then AK2 = AF2, that is, AM2 + MK2 = AH2 + HF2. But AH2>AM2. .. HF2 < MK2, that is, GF <LK. 11. Let A be any pt. on the circumference: ACB the diameter, AD any other chord. Then CBCD; .. AB = AC + CD > AD. And if AF is nearer to AB than AD, the two sides AC, CD are equal to the two sides AC, CF, but the angle ACF > angle ACD, .. base AF > AD [1. 24]. Make ACE ACF. Then AE AF [1.4]. But AF > AD. Hence two and only two chords from A can be drawn equal to one another. Page 173. 1. Since equal chords are equidistant from the centre, the locus is a circle, whose centre is the centre of the given circle and radius is equal to the distance of any of the chords from the centre. S 2. Let chords AB, CD cut in E. Let F be the centre. Draw FG, FH perp. to AB, CD. Then in the right-angled ▲ FEG, FEH, the hypotenuse EF is common and the angles GEF, HEF are equal. .. GF = HF [1. 26], .. the chords AB, CD are equidistant from the centre, and are therefore equal [ш. 14]. 3. Take fig. of preceding Ex. Then EG = EH. the halves of AB, CD [III. 3], are equal. mainder BE = whole or remainder CE. But BG, HD, .. the whole or re 4. With centre, A, on the Oce of the given O, describe a circle having radius equal to the required length, cutting the given circle again in B. From the centre of the given circle, draw a st. line perp. to the required direction and equal to the distance of the centre from AB. The chord drawn through the extremity of this st. line, par1. to the given direction, will be the chord required. 5. Let O be the centre: and AX, BY, OZ the perps. on PQ. Then OZ = the sum or difference of AX and BY, according as A and B are on the same or opposite sides of PQ [Ex. 18, p. 98]. .. the sum or difference of AX and BY = twice OZ = a constant. Page 175. Draw the 1. Let A be the given pt., and B the centre. chord CAD perp. to AB, and any other chord EAF through A. BG perp. to EF. Then the hypotenuse BA > BG. .. chord CD chord EF; that is, CD is the least chord through A. H. K. E. 4 |