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side of the lesser circle to O so that OG may be to OH as the radius GC to the radius HF (II. 55.): upon OG describe a semicircle cutting the circle ABC in B, so that OB and BG being joined may be perpendicular (15. Cor. 1.) to one another, and therefore OB a tangent at B (2.); and from H to OB (produced if necessary) draw the perpendicular HR. Then, because HR is parallel to GB (I. 14.) HR is to GB as OH to OG (II.30.Cor.2.), that is, as HF toGC; and, because GB is equal to GC, HR is equal to HF. Therefore R is a point in the circle DEF; and because ORH is a right angle, OB touches the circle DEF in R. Upon BR describe the semicircle BLMR cutting GH in the points L and M; and if PA is to be greater than PD, produce GH to K (II. 55.) so that GK may be to HK in the duplicate of the given ratio (as in the figure); but, if PA is to be less than PD, produce HG to K (II. 55.), so that GK may be to HK in the duplicate of the given ratio; take KN (II.51.) a mean proportional between KL and KM, and from the centre K with the radius KN describe the circle NPQ; this circle shall be the locus required.

For, if P be any point in the locus, and if the tangents PA and PD be drawn, and PL, PM joined, PA will be to PD in the ratio which is compounded of the ratios of PA2 to PM3, PM2 to PL2, and PL to PD. (II.def. 12.)

But the circle NPQ stands related to each of the circles ABC, DEF, with the corresponding points M, L, in the same manner in which KPQ is related to BCD in the preceding locus. For, with regard to ABC, because GB is perpendicular to BR, which is the diameter of the semicircle BLMR, GB touches the semicircle (2.) at B, and therefore (21.) GB2 is equal to GLX GM, that is, GL a third proportional to GM and GB, or GC; and KN was made a mean proportional between KL and KM. And, in the same manner, with respect to the other circle DEF, HM is a third proportional to HL and HF; and KN was made a KM. Therefore, by the last question, mean proportional between KL and PA is to PM as KG to KM; and PL is to PD as KL to KH. Also, because KL, KN and KM are proportionals, PM is to PL as KM to KL. (51.) Therefore the ratio of PA to PD is compounded of ratios which are the same with the ratios of KG to KM, KM to KL, and KL to KH, that is, it is the same with the ratio of KG to KH (II. 27.); and PA is to PD in the subduplicate ratio of KG to KH, that is in the given ratio. Therefore the circle NPQ is the locus required.

If the given ratio be the ratio of equality, the difference of the squares

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of PG, PH will be equal to the difference of the squares of GC and HF; and therefore the locus is a straight line (49.) cutting GH at right angles, and may be determined as in Prop. 49.

The first of the two loci we have thus discussed is manifestly the same which satisfies the condition that A and E being two given points, and ED a given square, PA shall be to PEED2 in a given ratio: and the second, the same which satisfies the condition that G and H being two given points, and GC and HF, two given squares, PG-GC shall be to PH-HF in a given ratio.

PROP. 52.

A point A being given within or without a circle BDE, and in every

chord MN which passes through it, a second point P being taken such that the chord produced may be divided by these two points and the circumference harmonically; it is required to find the locus of the points P.

Let C be the centre of the circle, and let C A, produced if necessary, meet the circle in B: take CF a third proportional to CA, CB, (II. 52.) and join FP, FM, FN. Then, because CF, CB, CA are proportionals, and that CE is equal to CB, the straight lines EA, EB, EF are in harmonical progression (II. 46.). And, because upon the mean EB, the circle EDB is described (51. Cor.), AM is to M Fas AN to NF. Therefore, alternando, (II. 19.) AM: AN:: MF: NF. 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.

passes through it, tangents being drawn intersecting in P; it is required to find the locus of the points P.

Let C be the centre of the circle, and let CA, produced if necessary, meet the circumference in B: take CF a third proportional to C A, C B: join PF, PC, CM, CN, and let P C cut M N in Q.

Then, because PM is equal to PN (2. Cor. 3.), and CM to CN, MN is bisected by PC at right angles (3. Cor. 3.). And, because CNP is a right angled triangle (2.), and that from the right angle N, a perpendicular N Q is drawn to the hypotenuse, the rectangle under CQ, CP is equal (I. 36. Cor. 2.) to the

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Consequently, as was shown in the like case in the demonstration of the last proposition, the angle AFP is a right angle, and the point P lies in a straight line drawn from the point F perpendicular to CF. It is easy to reverse the reasoning, and to show that every point in this straight line satisfies the given condition. Therefore this straight line is the locus required.

Cor. If the diameter of a circle, and the diameter produced, be divided in the same ratio, or, which is the same thing, (II.45 Cor.) if the diameter produced be divided harmonically, any chord which passes through one point of division shall be divided harmonically by the circumference and the perpendicular to the diameter which is drawn through the other point.

PROP. 53.

A point A being given within or without a circle BD E, and at the extremities of every chord MN which

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 equiangular with the triangle C Q A. Therefore CFP is a right angle, and the point P is in a straight line drawn through the point F perpendicular to CF. It is easy to reverse the reasoning, and to show that every point in this straight line satisfies the given condition. Therefore this straight line is the locus required.

Cor. If the diameter of a circle, and the diameter produced, be divided in the same ratio, or, which is the same thing, (II. 45 Cor.) if the diameter produced be divided harmonically, and if tangents be drawn at the extremities of any chord passing through one of the points of division, they shall intersect one another in the perpendicular to the diameter which is drawn through the other point.

SECTION 7.-Problems.

PROP. 54. Prob. 1. (Euc. iii. 30.) To bisect a given circular arc AC B. Let C be the required point of bisection: take D, the middle point of

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The practicability of a geometrical division of a circular arc into any number of equal parts, implies that of the angle at the centre (12.) into the same number of equal parts; and vice versa. It has already been stated that in the cases of 3, 5, &c., equal parts, the division of the angle cannot be effected by a plane construction; and the same is to be understood of the circular arc (I. 46. Scholium). We may observe that the problem of trisecting an arc has been put under the following form, which gives it an appearance at first of being much easier than upon examination it is found to be.

"From a given point A in the circumference of a given circle ABD B to draw a straight line APQ such that the part PQ, which

D

is intercepted between the circumference and a given diameter BD produced, shall be equal to the radius CA.

For, if this be done, the arc PD will be found, which is a third of the given arc AB; because, PQ being equal to PC, the angle PCQ is equal (1.6) to the angle PQC, and therefore, the angle CPA or CAP is equal to twice PCQ (I. 19.), and ACB, which (I. 19.) is equal to CAP and PQC together, is equal to three times PCQ, that is, the arc AB is equal to three times the arc PD (13.)

PROP. 55. Prob. 2. (Euc. iii. 1.) To find the centre of a given circular arc A CB.

Since the straight line which bisects a chord at right angles

passes through the centre of the circle, two such straight lines

A

will cut one another in the centre. Therefore, in the arc A CB take any point C; join A C, C B; and bisect A C, CB at right angles by the straight lines DE, FE: the point E in which they cut one another is the centre of the arc A CB. (See also Prop. 44.)

Cor. (Euc. iii. 25.) Hence, any arc of a circle being given, the circumference may be completed of which it is a part.

PROP. 56. Prob. 3. (Euc. iii. 17.) From a given point A, to draw a tangent to a given circle B D E.

B

1. If the point A be in the circumference of the circle, find the centre C D (55.), join C A, and from A draw A F perpendicular to CA. Then, because AF is drawn perpendicular to the radius at its extremity, it touches the circle (2.).

2. If A do not lie

in the circumference, let the line AB be assumed as the required tangent. Find the centre C, and join CA, C B. Then, be

D

E

E

3

A.

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B

E

Join CA, ca; and from c, the centre of the lesser circle, draw c E parallel to Aa to meet CA in E. Then, because C Aa (2. Cor. 1.) is a right angle, CEC is likewise a right angle (I. 14.), and CE will touch in the point E the circumference of a circle, described from the centre C, with the radius CE, which is equal to the difference of CA, EA, that is (I. 22.) to the difference of the radii C A, c a.

Therefore, reversely, from the centre C with a radius equal to the difference of the radii C A, ca, describe a circle, and from the point e draw a straight line (56.), touching this circle in the point E: join CE, and produce it to meet the circumference ABD in A; drawca parallel to CA, and join A a: Aa is the common tangent required. When the circles are equal, CA and camust be drawn, each of them, perpendicular to Cc. In this case the common tangent A a is evidently. parallel to C c.

2. If the points of contact are to lie

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PROP. 59. Prob. 6.

To describe a circle, which shall 1. (Euc. iv. 5.) pass through three given points not in the same straight line; or

2. pass through two given points, and touch a given straight line; or touch two given straight lines; or 3. pass through a given point, and

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

A

B

B

P

1. Let A, B, C be the three given points, and let the point P be assumed for the centre of the required circle. Then, because P is equidistant from A and B, it is in the straight line which bisects A B at right angles (44.); and for a similar reason, it is in the straight line which bisects BC at right angles. Therefore, reversely, to find the point P, bisect AB, BC at right angles (1.43. Cor.) by the straight lines DP, EP, which intersect one another in P: and, from the centre P with the radius PA, describe a circle; it shall pass through the points B, C, and shall be the circle required.

2. Let A, B be the two given points, and CD the given straight line. Suppose the circle to be described, and that it touches CD in P: also, let AB produced meet CD in C. Then, be

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each of which passes through the two given points A, B, and touches the given straight line CD, viz. one for a point P upon the right hand of C, and another for a point P upon the left hand of C.

If A B be parallel to CD, the point P will be in the straight line which bisects A B at right angles, (3. Cor. 1.) and being found accordingly, the circle may be described as before.

3. Let A be the given point, and BC, DE the given straight lines. Produce BC, DE to meet one another in F: join FA: draw FG bisecting the angle BFD (1.46.): in FG take any point G, and from G draw GE (I. 45.) perpendicular to FD: from the centre G with the radius GE describe a circle cutting FA in H: join HG, and through A draw A P parallel to HG to meet FG (produced, if necessary), in P (I. 48.): the circle described from

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the centre P with the radius P A shall be the circle required.

For, if PD be drawn from the point P perpendicular to FD, PD will be parallel to GE (I. 14.), and, therefore, (II. 30. Cor. 2.) GE will be to PD as GF to PF: but, because G H is parallel to PA, GF is to PF as GH to PA: therefore (II. 12.) GE is to PD as GH to PA: and in this proportion the first term GE is equal to the third GH; therefore also (II. 18.) PD is equal to PA. Therefore, the circle which is described from the centre P with the radius PA passes through the point D; and it touches the line D E in that point, because PD is at right angles to DE (2.): therefore (45.) it also touches

the line B C.*

The case in which BC is parallel to DE differs from the foregoing in this only, that FG must now be drawn parallel to BC or DE, and bisecting the distance between them (See 45.).

When the point A is in FG, or in FG produced, the solution here given must be modified by joining HE, drawing AD parallel to H E, and erecting DP perpendicular to FD; which gives the centre P as before. When A is in one of the given lines, as BC, the solution takes a still more simple form.

In both cases we may observe that there are two circles which satisfy the given conditions, corresponding to the two points in which the circle which is described from the centre G with the radius GE cuts the line FA.

4. Let A B, BC, CD be the three given straight lines, of which not more than two are parallel, and let these two be cut by the third in the points B, C.

A

B

Assume the point P for the centre of the circle. Then, because P is equidistant from AB, BC, it is in the straight line BP which bisects the angle ABC (45.); and, for a similar reason, it is in the line PC which bisects the angle B CD. Therefore, reversely, to find P, bisect the angles at B and C by straight lines meeting in P: from P draw PQ perpendicular to A B (I. 45.), and from the centre P with the radius PQ describe a circle: it shall be the circle required.

If AB, CD 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 paralshall satisfy this condition, viz. one lel, four circles may be described which given lines, and three others touching within the triangle included by the the sides of that triangle externally.

Scholium.

The problem of describing a circle about a given triangle (Euc. iv. 5.) belongs to the first case, that of inscribing a circle within a given triangle (Euc. iv. 4.) to the last case of this proposition. The second and third cases are modified by supposing a point and a tangent passing through it to be of the data. Thus, the second becomes "to describe. a circle, which shall pass through two given points, and touch a given straight line in one of those points," and the third "to describe a circle which shall touch two given straight lines, and one of them in a given point." The modified solutions corresponding are too simple to detain us here: that of the first occurs in prob. 7.

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Instead of touching one, two, or three given straight lines as in the problem we have just considered, it may be required to describe a circle which shall

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