The 32d proposition of the preceding book, when considered in a particular manner, affords another instance of a porism; as it appears that if a circle and a point D or E be given, another point E or D may be found, such that any circle whatever, described through D and E will bisect the circumference of the given circle; and this may be regarded as the indeterminate case of the problem, in which it is required, through two given points to describe a circle bisecting the circumference of another given circle;-a problem which is always determinate, except when the points are situated in the manner supposed in the proposition referred to.

PROPOSITIONS.

PROP. I. PROB.*-Through two given points, to describe a circle touching a straight line given in position.

Through two given points A, B, let it be required to describe a circle, touching a given straight line CD.

Join AB. Then, if CD be parallel to AB, bisect AB by the perpendicular EF, and (III. 1, schol.) through A, F, B describe a circle: it is the circle required. For (III. 1, cor.) EF passes through the centre, and (I. 29) is perpendicular to CD; therefore (III. 16) the circle touches CD; and it passes through A, B.

F

D

C

F

E

F

A

E

B

D

But if CD, BA be not parallel, let them meet in E: between AE, BE find a mean proportional, and make EF equal to it: through B, A, F describe a circle: it is the one required. For (const. and VI. 17) the rectangle BE.EA is equal to the square of EF, and therefore (III. 37) EF touches the circle; and the circle passes through the given points.

B

Schol. 1. In this latter case, since EF may be taken on either side of E, there are two circles that answer the conditions of the problem; and these are evidently unequal, unless BAE be perpendicular to CD.

Schol. 2. If one of the points A be in the given line, the pre

*This is one of the problems known under the name of tangencies. The general problem of the tangencies, as understood by the ancients, is as follows: Of three points, three straight lines, and three circles of given radii, any three being given in position; it is required to describe a circle passing through the points, and touching the straight lines and circles. This general problem comprehends ten subordinate ones, the data of which are as follows: (1.) three points; (2.) two points and a straight line; (3.) two points and a circle; (4.) a point and two straight lines; (5.) a point, a straight line, and a circle; (6.) a point and two circles; (7.) three straight lines; (8.) two straight lines and a circle; (9.) a straight line and two circles; and (10.) three circles. The first and seventh of these are the fifth and fourth propositions of the fourth book of Euclid: the rest are given here.

If a circle be continually diminished, it may be regarded as becoming ultimately

ceding solution fails, as it does not give a third point in the required circumference. In that case two straight lines, one bisecting AB perpendicularly, and the other drawn through A perpendicular to CD, will intersect each other in the centre of the required circle. The proof is plain from the fourth proposition of the first book, and the sixteenth of the third book.

PROP. II. PROB.-Through two given points, to describe a circle touching a given circle.

Through the given points A, B, let it be required to describe a circle touching a given circle CDE.

Join AB. Then, if A, B be equally distant from the centre of the given circle, bisect AB by a perpendicular, cutting the given circle in two points, and it is plain that a circle described through either of these points, and through A, B, will touch the given circle.

But if A, B be not equally distant from the centre, take any point C in the circumference of CDE, and describe a circle through A, B, C: if this touch the given circle, it is a circle such as is required. But if it do not, let it cut CDE again in D, and draw CDF meeting AB pro

E

A

duced in F: draw (III. 17) FE touching the given circle in E: through ABE describe a circle: it is the one required.

Join AD, BC. Then (III. 21) the angles A, C are equal, as they stand on the same arc BD: also AFC is common to the two triangles AFD, CFB. Therefore (VI. 4, cor.) AF: FD :: CF : FB; and hence (VI. 16) the rectangles AF.FB, CF.FD are equal. But (III. 36) CF.FD is equal to the square of EF: wherefore also AF.FB is equal to the square of EF. ABE therefore (III. 37) touches the straight line FE; and consequently it touches the circle CDE.

Schol. If one of the points A be on the given circumference, the preceding method fails, as it does not give a third point in the required circumference. In that case the construction is effected simply by drawing a radius to A, and producing it, if necessary, to meet a perpendicular bisecting AB: their point of intersection is evidently the centre of the required circle.

B

In this case there is but one circle that answers the conditions of the problem. When the points are both without the circle, or both within it, there are two circles; but if one be without and the other within, the problem is manifestly impossible.

a point. By being continually enlarged, on the contrary, it may have its curvature so much diminished, that any portion of its circumference may be made to differ

Cor. 1. Hence in a given straight line AB, a point may be found, such that the difference of its distances from two given points C, D, may be equal to a given straight line. Join CD, and from D as centre, with a radius, DE equal to the given difference, describe a circle; draw CFG perpendicular to AB, and make FG equal to FC; through C, G, describe a circle touching the other circle; join B, D, the centres of the two circles, and draw BC: BC, BD are evidently the required lines.

E

Cor. 2. In the same manner, if a circle were described from D as centre, with the sum of two given lines as radius, and a circle were described through C and G touching that circle, straight lines drawn from the centre of that circle to C and D, would be equal to the given sum.

PROP. III. PROB.-Through a given point, to describe a circle touching two given straight lines.

Through a given point A, let it be required to describe a circle touching two given straight lines, BC, DE.

First, let BC, DE not be parallel, but let them meet in B: draw BF bisecting the angle CBE and through A, draw LFAH, perpendicular to BF, cutting DE in H, and BC in L; and make FG equal to FA. Take HK a mean proportional between AH, HG; through K, A, G describe a circle: it is the one required.

N

C

G

For, in the triangles BFL, BFH, the angles at B and Fare equal, and BF common therefore (I. 26) FL, FH are equal; and (I.ax. 3) LG, AH are equal. But (const. and VI. 17) the rectangle GH.HA is equal to the square of HK; and therefore (III. 37) DE touches the circle. Now (III. 1, cor.) since FA, FG are equal, and FB perpendicular to AG, the centre of the circle KAG is in FB: let it be M, and join MK; draw also MN perpendicular to BC. Then (I. 26) in the triangles MBK, MBN, MN is equal to MK, and the circle will pass through N, and (III. 16) will touch BC, since BNM is a right angle.

D

M

A

H

K E

Secondly, let BC, DE be parallel, and A be between them.

in as small a degree as we please from a straight line. Viewing the subject in this light, we may regard the first nine of the problems, now mentioned, as comprehended in the tenth. Thus, we shall have the first, by supposing the circles to become infinitely small; the seventh, by supposing them infinitely great; the fifth, by taking one of them infinitely small, one infinitely great, and one as a circle of finite magnitude: and so on with regard to the others. These views of the subject tend to illustrate it; but they do not assist in the solution of the problems.

с

Through A draw LAH perpendicular to the parallels: bisect HL in F, and make FG equal to FA: take HK a mean proportional between GH, HA; and a circle described through GAK is the one required. The proof is easily derived from that of the preceding

case.

B

L

G

F

A

L K

H

K

Schol. If the given point A be in one of the given lines DE, the foregoing solution fails. In that case, find F as before, and a perpendicular through A to DE will intersect BF in the centre of the required circle.

In this case, there may be two circles, one touching DE, BC, and the other DE and the continuation of BC, through B, unless BC, DE be parallel, when there is but one circle. When A is not on BC or DE, there may obviously be two circles, since HK may be taken on either side of H.

PROP. IV. PROB.-Through a given point, to describe a circle, touching a given straight line, and a given circle.

Through a given point A, let it be required to describe a circle touching a given straight line BC, and a given circle DEF.

Through G, the centre of DEF, draw DEB perpendicular to BC, and join AD; to AD, DB, DE, take DH a fourth proportional; through A, H, describe (APP. II. 1) the circle HAC touching BC; HAC is the circle required.

H

F

Join the centre L, and the point of contact C: then (III. 18) LCB is a right angle, and LC, DB are parallel. Draw CD, cutting the circle DEF in F, and join EF. Then the triangles DBC, DFE are similar, and therefore CD: DB:: DE: DF; whence (VI. 16) the rectangle CD.DF is equal to BD.DE: but (const. and VI. 16) AD.DH is equal to BD.DE: therefore CD.DF is equal to AD.DH. Hence (APP. I. 5, schol.) the point F is also in the circumference of AHC. Join FG, FL. Then (I. 29) the angles GDF, LCF are equal; wherefore DFG, CFL,

E

A

which (I. 5) are respectively equal to these, are equal to one another and DFC being a straight line, GFL (I. 14, cor.) is also a straight line. Hence (APP. I. 2, cor.) the circles touch one another in F; for they meet in F, and the straight line joining their centres passes through that point.

Schol. This solution fails, if A be either at C, any point in BC, or at F, any point in the circumference of DEF. The construction is then effected simply by drawing DFC, GFL, and the perpendicular CL; as CL cuts GL in the centre of the required circle. It may be farther remarked, that, when the given point, straight line, and circle, are situated as in the diagram here employed, there will be two circles obtained by the construction given above, each

of which will be touched externally by the given circle. Besides these, two others, which will be touched internally by the given circle, will be obtained, by joining AE instead of AD, and making the variations thence arising. In making the construction in this manner, H is taken in the continuation of the line joining A with the extremity of the diameter. The variations in the proof present no difficulty.*

PROP. V. PROB.-Through a given point, to describe a circle touching two given circles.

Through a given point A, let it be required to describe a circle touching two given circles, BCD, EFG.

Through the centres H, K, draw BDEGL, and (VI. 10, schol.) find the point L, such that HL is to KL, as the radius of the greater circle to the radius of the less join AL, and (VI. 12)

:

take ML such that AL BL :: GL: LM: then (APP. II. 2) through A and M describe the circle ACM, touching BCD: it is the circle required.

Through the point of contact C, draw LCN: join HN, HC, and let KF, KO be drawn parallel to them, and meet LN in F, 0. Then (VI. 4 and alternately) HL: KL:: HN: KF, and HL: KL :: HC KO; but (const.) HL is to KL, as the radius of BCD to the radius of EFG; therefore KF, KO are radii of EFG, and LN cuts EFG in F and O. Join BC, FG. Then the triangles LBC, LFG have an angle common, and the angles LBC, LFG equal, because (III. 20) they are halves of the equal angles DHC, GKO; and therefore BL LC :: FL: LG, and (Vl. 16) the

* In this problem, the straight line may fall without the circle, may cut it, or may touch it: the point may be without the circle, within it, or in its circumference; or it may be in the given straight line, or on either side of it: and it will be an interesting exercise for the student, in this and many other problems, to consider the variations arising in the solution, from such changes in the relations of the data, and to determine what relations make the solution possible, and what render it impossible. It may also be remarked, that in many problems, there will be slight variations in the proofs of different solutions of the same problem, even when there is no change in the method of solution; such as in the present instance, when the required circle is touched externally, and when internally. Thus, while in one case angles may coincide, in another the corresponding ones may be vertically opposite; and the reference may sometimes be to the third, and sometimes to the fourth proposition of the first book of this Appendix. It is, in general, unnecessary to point out these variations, as, though they merit the attention of the student, they occasion no difficuity.