chord M N which passes through it, a passes through it, tangents being drawn second point P being taken such that the intersecting in P; it is required to find chord produced may be divided by these the locus of the points P. two points and the circumference har- Let C be the centre of the circle, and monically; it is required to find the let C A, produced if necessary, meet the locus of the points P.

circumference in B: take CF a third Let C be the centre of the circle, and proportional to C A, C B :join PF, PC, let C A, produced if necessary, meet the CM, CN, and let PC cut M N in Q. circle in B: take CF a third propor- Then, because PM is equal to PN tional to CA, CB, (II. 52.) and join FP, (2. Cor. 3.), and CM to CN, MN is biFM, FN. Then, because CF, CB, CA sected by PC at right angles (3. Cor. 3.). are proportionals, and that CE is equal And, because CNP is a right angled to CB, the straight lines EA, EB, EF triangle (2.), and that from the right are in harmonical progression (11. 46.). angle N, a perpendicular N Q is drawn And, because upon the mean EB, the to the hypotenuse, the rectangle under circle EDB is described (51. Cor.), AM CQ, CP is equal (I. 36. Cor. 2.) to the is to M Fas AN to NF. Therefore, alternando, (II. 19.) A M:AN::MF:N F. But by the supposition that P is a point of the required locus, MP: PN:: AM : AN, that is, :: MF: NF: therefore, in the triangle FMN, both the base M N and the base produced are divided in the ratio of the sides.

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square of CN, that is, to the square of CB, or to the rectangle CA, CF (II. 38. Cor. 1.). Therefore (II. 38.) CQ is to CA as CF to CP,

and (II. 32.) the triangle CFP is equi; Consequently, as was "shown in the angular with the triangle C Q A. There like case in the demonstration of the last fore CFP is a right angle, and the

point P is in a straight line drawn proposition, the angle AFP is a right angle

, and the point P lies in a straight line through the point F perpendicular to drawn from the point F perpendicular to and to show that every point in this

CF. It is easy to reverse the reasoning; CF. It is easy to reverse the reasoning, straight line satisfies the given condition. and to show that every point in this Therefore this straight line is the locus straight line satisfies the given condition.

required. Therefore this straight line is the locus

Cor. If the diameter of a circle, and required.

Cor. If the diameter of a circle, and the diameter produced, be divided in the the diameter produced, be divided in the (II. 45 Cor.) if the diameter produced be

same ratio, or, which is the same thing, same ràtio, or, which is the same thing, divided harmonically, and if tangents be (II.45 Cor.) if the diameter produced be drawn at the extremities of any chord divided harmonically, any chord which passes through one point of division shall passing through one

of the points of dibe divided harmonically by the circum- in the perpendicular to the diameter

vision, they shall intersect one another ference and the perpendicular to the which is drawn through the other point. diameter which is drawn through the other point.

SECTION 7.--Problems.
PROP. 53.

PROP. 54. Prob. 1. (Euc. iii. 30.) A point A being given within or To bisect a given circular arc AC B. without a circle B D E, and at the ex- Let C be the required point of bitremities of every chord M N which section: take D. the middle point of





AB; and join CA, CB, CD. Then, Since the straight because the arc CA

line which bisects a is equal to the arc

chord at right angles CB, the chord CA (12.

passes through the Cor. 1.) is equal to the

centre of the circle, chord C B : and, be

two such straight lines cause the triangles CDA, CDB. have will cut one another in the centre. the three sides of the one equal to the Therefore, in the arc A CB take any three sides of the other, each to each, point C; join A C, C B; and bisect A C, the angle CDA is equal to CDB: there. C B at right angles by the straight lines fore, the line CD bisects AB at right DE, FE: the point E in which they angles. Therefore reversely, bisect the cut one another is the centre of the arc chord AB at right angles by the straight A CB. (See also Prop. 44.) line CD, and the given arc ACB will be Cor. (Euc. iii. 25.) Hence, any arc of bisected in C.

a circle being given, the circumference Therefore, &c.

may be completed of which it is a part. Cor. Hence, a given arc may be divided into 4, 8, 16, &c. equal parts. PROP. 56. Prob. 3. (Euc. iii. 17.) Scholium.

From a given point A, to draw a tanThe practicability of a geometrical gent to a given circle B D E. division of a circular arc into any num- 1. If the point A ber of equal parts, implies that of the be in the circumfeangle at the centre (12.) into the rence of the circle, same number of equal parts; and vice find the centre Ć versâ. It has already been stated that (55.), join CA, in the cases of 3, 5, &c., equal parts, and from A draw the division of the angle cannot be ef- A F perpendicular fected by a plane construction; and the to C Å. Then, because AF is drawn same is to be understood of the circular perpendicular to the radius at its extrearc (I. 46. Scholium). We may observe mity, it touches the circle (2.). that the problem of trisecting an arc 2. If A do not lie has been put under the following form, in the circumference, which gives it an appearance at first of let the line A B be being much easier than upon examina- assumed as the retion it is found to be.

quired tangent. Find “ From a given

the centre C, and join point A in the cir.

CA, CB. Then, becumference of a

cause AB is a tangent, the angle CBA is a given circle ABD B

right angle (2. Cor. 1.), and the point B to draw a straight

lies in the circumference of a circle of line APQ such that

which A C is diameter (15. Cor. 3.) the part PQ, which

Reversely, therefore, upon A C as a diais intercepted between the circumference meter describe a circle cutting the given and a given diameter BD produced, shall circle in B, and join AB: A B is the be equal to the radius CA.”

tangent required. For, if this be done, the arc PD will We may observe that in this case be found, which is a third of the given there are two points of intersection B,

The same arc AB; because, PQ being equal to and therefore two tangents. PC, the angle PCQ is equal (1.6) to the may be said, indeed, of the former case; angle PQČ, and therefore, the angle but there the two touch in the same CPA or CAP is equal to twice PCQ point, and are parts of the same straight (I. 19.), and ACB, which (I. 19.) is equal line. to CAP and PQC together, is equal to three times PCQ, that is, the arc

PROP. 57. Prob. 4. AB is equal to three times the arc PD

To a given circle A B D to draw a (13.)

tanyent, which shall be parallel to a Prop. 55. Prob. 2. (Euc. iii. 1.) given straight line E F. To find the centre of a given circular Find the centre C (55.); from C draw arc A CB.

CE perpendicular to EF (I. 45.), and










let CE, produced if

upon opposite sides of the line joining necessary, cut the cir

the centres, the like reasoning will lead cumference in P.

to a construction which differs from Then, if PQ be drawn

that of the first case in this only, that parallel to EF, it will

CE is equal to the sum of the radii. be at right angles to

We may remark that in each of the the radius CP (I.

cases there are two tangents c E, and 14.); and therefore

two common tangents parallel to them (2.) will be the tan

respectively. gent required.

PROP. 59. Prob. 6.
PROP. 58. Prob. 5.
To draw a straight line, which shall

To describe a circle, which shall touch each of two given circles A BD, 1. (Euc. iv. 5.) pass through three abd.

given points not in the same straight Let A a be assumed as the common

line; or tangent required: and first, let it touch

2. pass through two given points, the circles in the points A, a which lie and touch a given straight line; or upon the same side of the line joining touch two given straight lines ; or

3. pass through a given point, and their centres C, c.

4. (Euc. iv. 4.) touch three given straight lines not parallels.

1. Let A, B, C be the three given

points, and let the point P be assumed B

for the centre of the required circle.
Then, because P is
equidistant from A

and B, it is in the Join C A,ca; and from c, the centre straight line which of the lesser circle, draw cE parallel bisects A B at right to A a to meet CA in E. Then, be- angles (44.) ; and cause C Aa (2. Cor. 1.) is a right angle, for a similar reason, CEc is likewise a right angle (I. 14.), it is in the straight and C E will touch in the point E the line which bisects BC at right angles, circumference of a circle, described Therefore, reversely, to find the point P, from the centre C, with the radius CE, bisect AB, BC at right angles(I.43. Cor.) which is equal to the difference of CA, by the straight lines DP, EP, which inEA, that is (1. 22.) to the difference of tersect one another in P: and, from the the radii CA, C a.

centre P with the radius PA, describe Therefore, reversely, from the centre C with a radius equal to the difference B, C, and shall be the circle required.

a circle; it shall pass through the points of the radii CA, ca, describe a circle,

2. Let A, B be the two given points, and from the point c draw a straight and CD the given straight line. Supline (56.), touching this circle in the point E : join CE, and produce it to meet it touches C D in P: also, let AB

pose the circle to be described, and that the circumference ABD in A; draw ca parallel to C A, and join A a: 'A a is the produced meet CD in C. Then, becommon tangent required. When the circles are equal, CA and camust be drawn, each of them, perpendicular to C c. this case the common tangent A a is evidently parallel to C C. 2. If the points of contact are to lie

cause C P touches, and C A cuts the circle, the square of C P is equal (21.) to the rectangle under CA, CB. Therefore, reversely, take C P a mean proportional between CA, C B, and describe a circle through the points A, B, P; it shall be the circle required.

We may remark that in this case, there may be described two circles,

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each of which passes through the two In both cases we may observe that given points A, B, and touches the there are two circles which satisfy the given straight line C D, viz. one for a given conditions, corresponding to the point P upon the right hand of C, and two points in which the circle which is another for a point P upon the left described from the centre G with the rahand of C.

dius G E cuts the line FA. If A B be parallel to C D, the point 4. Let A B, BC, P will be in the straight line which bi- CD be the three sects A B at right angles, (3. Cor. 1.) given straight lines, and being found accordingly, the circle of which not more may be described as before.

than two are paral3. Let A be the given point, and BC, lel, and let these DE the given straight lines. Produce two be cut by the BC, DE to meet one another in F: third in the points join FA: draw F G bisecting the angle B, C. BFD (1.46.): in FG take any point Assume the point P for the centre of G, and from G draw G E (1. 45.) per- the circle. Then, because P is equipendicular to FD: from the centre G distant from AB, BC, it is in the with the radius GE describe a circle straight line B P which bisects the ancutting FA in H: join HG, and gle ABC (45.); and, for a similar reason, through A draw A P parallel to HG it is in the line PĆ which bisects the to meet FG (produced, if necessary), in angle BCD. Therefore, reversely, to P (I. 48.): the circle described "from find P, bisect the angles at B and © by

straight lines meeting in P: from P draw P Q perpendicular to A B (1. 45.), and from the centre P with the radius PQ describe a circle: it shall be the circle required.

If AB, C D are parallel, two circles (and two only) can be described, each touching the three given lines : but

if no two of the straight lines be paralthe centre P with the radius P A shall shall satisfy this condition, viz. one

lel, four circles may be described which be the circle required. For, if P D be drawn from the point given lines, and three others touching

within the triangle included by the P perpendicular to FD, PD will be the sides of that triangle externally. parallel to G E (I. 14.), and, therefore, (II. 30. Cor. 2.) GE will be to PD as GF

Scholium. to PF: but, because G H is parallel to PA, G F is to PF as GH to PA: The problem of describing a circle therefore (II. 12.) G E is to PD as GH about a given triangle (Euc. iv. 5.) beto PA: and in this proportion the first longs to the first case, that of inscribterm GE is equal to the third GH; ing a circle within a given triangle therefore also (II. 18.) PD is equal tó (Euc. iv. 4.) to the last case of this proPA. Therefore, the circle which is de- position. The second and third cases scribed from the centre P with the ra- are modified by supposing a point and a dius P A passes through the point D; tangent passing through it to be of the and it touches the line D E' in that data. Thus, the second becomes “to point, because PD is at right angles to describe a circle, which shall pass D E (2.): therefore (45.) it also touches through two given points, and touch a the line B C.*

given straight line in one of those The case in which B C is parallel to points,” and the third “to describe a DE differs from the foregoing in this circle which shall touch two given only, that FG must now be drawn straight lines, and one of them in a parallel to B C or DE, and bisecting given point." The modified solutions The distance between them (See 45.). corresponding are too simple to detain

us here: that of the first occurs in

prob. 7. * When the point A is in F G, or in FG produced, the solution here given must be modified by joining

Instead of touching one, two, or three 11 E, drawing A D parallel to H E, and erecting given straight lines as in the problem DP perpendicular to FD; which gives the centre P we have just considered, it may be re

When A is in one of the given lines, as BC, the solution takes a still more simple form.

quired to describe a circle which shall


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touch one, two, or three given circles, after a similar manner; viz. by describor which shall touch both a straighting a circle which shall touch the given line and a circle, or two straight lines straight line in the given point, and likeand a circle, or a straight line and two wise pass through a point assumed in circles, and at the same time pass the circumference of the circle, and then through one or two given points, as proceeding as in the proposition. the other data may happen to admit. 2. To describe a circle which shall The six new problems of contact which pass through a given point A, and are thus suggested are too remarkable touch two given circles B C D E F G. to be passed over without further no- Suppose that the required circle is tice; they are accordingly here sub- described, and that it touches the given joined.

circles in the points D, F respectively: 1. To describe a circle which shall join DF, and since the straight line pass through two given points A, B, DF cannot touch the given circles in and touch a given circle C D E. the points D, F, (because then it would

In the circumference C D E take any point C, and describe (59.) a circle which may pass through the three points A, B, C; then, if this circle

M meet the circumference CDE in no other point, it is the circle re

A quired : but if it do, let that other

touch the required circle ADF in the point be D: join

same points, which (1. Cor. 2.) is absurd) AB, CD, and

let it be produced both ways to meet the let them be pro

circumferences a second time in the duced to meet in

points C, G respectively: take H, K, the F: from F draw

centres of the circles B CD, E FG, and FG, touching the

join HK, HD, KF, KG: then, because circle CDE (56.):

the circle ADF touches BCD in D, let G be the point

and EFG in F, the radii HD; KF of contact, and

produced (8. Cor. 1.) will meet in its describe a circle (59.) passing through centre, L; and, because KGF, LFD the three points A, B, G. Then, be- are isosceles triangles, the angle KGF cause the chord C D of the circle is (I. 6.) equal to KFG, that is (I. 3.) CDE meets the tangent GF in F to L FD, that is again (I. 6.) to IDF (21.), the square of G F is equal to the or (I. 3.) HDC: therefore (I. 15.) KG rectangle CF, FD, that is, to the is parallel to HD, and, consequently, if rectangle AF, FB (20.): therefore the circles B C D E F G be equal, (21. Cor.) GF touches also the circle H K, D G will be parallel (I. 21.), or if ABG, and, consequently, the circle one of them, as B CD, be greater ABG touches the circle Č D E (2. Cor. than the other, H K and D G will meet, 2 and 9.), and is the circle required. if produced in some point M. In the

If A B and C D be parallels (which latter case, draw MB touching the will be the case when the line which bi- circle B C D in B (56.), join H B, and sects A B at right angles passes through from K draw KI (I. 45.) perpendicular the centre of the circle C D E), a tan- to MB, and therefore (I. 14.) parallel gent FG is to be drawn parallel to to HB: then, because KI, HB are A B or CD (57.), and the circle ABG, parallel, KI : HB::KM : HM (II. 30. being then described as before, will be Cor. 2.) that is, :: KG:HD, because the circle required.

KG, HD are parallel: but HB is It is evident, in each case, that there equal to HD; therefore (II. 18.) KI is are two tangents FG, and two circles equal to KG, and I is a point in the cirABG corresponding to them, one cumference of the circle EFG, and (2.) touching the given circle externally, MB touches the circle EFG in the the other internally.

point I. Now, MI: MB :: MK: Cor. The problem which requires a MH (II. 29.), that is, :: MG: MD; circle to be described which shall touch therefore, alternando (II. 19.), MI: MG a given straight line in a given point, ::MB : MD: but, because MI touches and also a given circle, may be solved the circle E FG, the square of M I is


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