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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 F as 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 PC 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
B I C
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 QA. ThereConsequently, as was shown in the fore CFP is a right angle, and the like case in the demonstration of the last point P is in a straight line drawn proposition, the angle AFP is a right an- through the point F perpendicular to gle, and the point P lies in a straight line CF. It is easy to reverse the reasoning, drawn from the point F perpendicular to and to show that every point in this CF. It is easy to reverse the reasoning, and to show that every point in this straight line satisfies the given condition. straight line satisfies the given condition. Therefore this straight line is the locus required. 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.
A point A being given within or without a circle BD E, and at the extremities of every chord MN which
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 which is drawn through the other point. in the perpendicular to the diameter
PROP. 54. Prob. 1. (Euc. iii. 30.)
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 cir cumference of a given circle ABD B to draw a straight line APQ such that the part PQ, which
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 (I. 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 C B.
Since the straight line which bisects a chord at right angles passes through the A centre of the circle, two such straight lines
will cut one another in the centre. Therefore, in the arc ACB 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 DE.
1. If the point A be in the circumference of the circle, find the centre C (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
cause AB is a tangent, the angle CBA is a right angle (2. Cor. 1.), and the point B lies in the circumference of a circle of which A C is diameter (15. Cor. 3.) Reversely, therefore, upon A C as a diameter describe a circle cutting the given circle in B, and join AB: A B is the tangent required.
We may observe that in this case there are two points of intersection B, and therefore two tangents. may be said, indeed, of the former case; but there the two touch in the same point, and are parts of the same straight line.
Join CA, ca; and from c, the centre of the lesser circle, draw c E parallel to A a to meet CA in E. Then, because CA a (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 C E, 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 c draw a straight line (56.), touching this circle in the point E: join CE, and produce it to meet the circumference ABD in A; draw ca parallel to CA, and join A a: A a 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
upon opposite sides of the line joining the centres, the like reasoning will lead to a construction which differs from that of the first case in this only, that CE is equal to the sum of the radii.
We may remark that in each of the cases there are two tangents c E, and two common tangents parallel to them respectively.
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.
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 (I.43. Cor.) tersect one another in P: and, from the by the straight lines DP, EP, which incentre 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 C D in P: also, let A B produced meet CD in C. Then, be
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 (I.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
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 P F: 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
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.
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.
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
touch one, two, or three given circles, or which shall touch both a straight line and a circle, or two straight lines and a circle, or a straight line and two circles, and at the same time pass through one or two given points, as the other data may happen to admit. The six new problems of contact which are thus suggested are too remarkable to be passed over without further notice; they are accordingly here subjoined.
1. To describe a circle which shall pass through two given points A, B, and touch a given circle C D E.
after a similar manner; viz. by describing a circle which shall touch the given straight line in the given point, and likewise pass through a point assumed in the circumference of the circle, and then proceeding as in the proposition.
2. To describe a circle which shall pass through a given point A, and touch two given circles BC D, E F G.
Suppose that the required circle is described, and that it touches the given circles in the points D, F respectively: join D F, and since the straight line DF cannot touch the given circles in the points D, F, (because then it would
In the circumference CDE take any point C, and describe (59.) a circle which may pass through the three points A, B, C; then, if this meet the circumference CDE in no other point, it is the circle required: but if it do, let that other point be D join A B, CD, and let them be produced to meet in F: from F draw FG, touching the circle CDE (56.): let G be the point of contact, and F describe a circle (59.) passing through the three points A, B, G. Then, because the chord CD of the circle CDE meets the tangent GF in F (21.), the square of G F is equal to the rectangle CF, FD, that is, to the rectangle AF, FB (20.): therefore (21. Cor.) GF touches also the circle AB G, and, consequently, the circle ABG touches the circle CDE (2. Cor. 2 and 9.), and is the circle required.
If A B and C D be parallels (which will be the case when the line which bisects A B at right angles passes through the centre of the circle CDE), a tangent FG is to be drawn parallel to A B or CD (57.), and the circle A B G, being then described as before, will be the circle required.
It is evident, in each case, that there are two tangents FG, and two circles ABG corresponding to them, one touching the given circle externally, the other internally.
Cor. The problem which requires a circle to be described which shall touch a given straight line in a given point, and also a given circle, may be solved
touch the required circle ADF in the same points, which (1. Cor. 2.) is absurd) let it be produced both ways to meet the circumferences a second time in the points C, G respectively: take H, K, the centres of the circles B CD, E F G, and join H K, HD, K F, KG: then, because the circle ADF touches BCD in D, and EFG in F, the radii HD, KF produced (8. Cor. 1.) will meet in its centre, L; and, because KGF, LFD are isosceles triangles, the angle KG F is (I. 6.) equal to KF G, that is (I. 3.) to LFD, that is again (I. 6.) to LD F or (I. 3.) H D C: therefore (I. 15.) K G is parallel to HD, and, consequently, if the circles BCD, EFG be equal, H K, D G will be parallel (I. 21.), or if one of them, as B C D, be greater than the other, H K and D G will meet, if produced in some point M. In the latter case, draw MB touching the circle BCD in B (56.), join H B, and from K draw KI (I. 45.) perpendicular to MB, and therefore (I. 14.) parallel to HB: then, because KI, HB are parallel, KI: HB::KM: HM (II.30. Cor. 2.) that is, KG: HD, because KG, HD are parallel: but HB is equal to HD; therefore (II. 18.) KI is equal to K G, and I is a point in the circumference of the circle E FG, and (2.) MB touches the circle EFG in the point I. Now, MI: MB :: MK: M H (II. 29.), that is, :: MG: MD; therefore, alternando (II.19.), MI: MG :: MB: MD: but, because MI touches the circle EFG, the square of MI is