в H с E E 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. 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. с Prop. III. PROB. - Through a given point, to describe a circle touching two given straight lines. Through a given point A, let it 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. For, in the triangles BFL, BFH, the angles at B and F are 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. Secondly, let BC, DE be parallel, and A be between them. G N M м A D H KE 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. B с L 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 F case. L K H 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. F E 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. 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, 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 per-pendicular 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 L B 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 EI and LN cuts EFG in F and 0. 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 (V1. 16) thé * 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 difflcuity. H с K rectangle BL.LG is equal to CL.LF. But (const. and VI. 16) BL.LG is equal to AL.LM: therefore the rectaugles CL.LE, AL.LM are equal, and (APP. I. 5, cor.) F is in the circumference of ACM. Produce HC, KF to meet in P. Then the angles PCF, PFC are equal, being equal to HCN, KFO, which are each equal to HNC: PC therefore is equal to PF. Now HP passes through the centre of ACM, for (III. 12) the straight line joining H with the centre of ACM passes through C. Hence, since CP, PF are equal, P is evidently the centre; * and therefore (APP. I. 2, cor.) ACM touches EFG in F. Prop. VI. Prob.—To describe a circle touching two given straight lines and a given circle. Let AB, CD be two straight lines given in position, and EF a given circle: it is required to describe a circle touching AB, CD, EF. Draw HK, LM parallel to CD, AB, at distances from them equal each to the radius GF: find (APP. II. 3) N, the centre of a circle passing through G, and touching HK, LM: draw GFN, and from N as centre, with NF as radius, describe the circle FDB: it is the circle required. Draw the perpendiculars NK, NM: these are each equal to GN; and GF, KD, MB being equal, NF, ND, NB are also equal, and the circle must pass through B, D: it also touches AB, CD, because the angles at B and D are right angles. PROP. VII. PROB.—To describe a circle touching two given circles, and a given straight line. Let AB, CD be the given circles, and EF the given straight line: it is required to describe a circle touching AB, CD, EF. Draw LM parallel to EF, and at a distance from it equal to BH; draw the radii CG, BH, and make CK equal to BH; with GK as radius, and G as centre, describe a circle KN; find (APP. II. 4)the centre of a circle touching KN and LM, and passing through H: it would be shown in nearly the same manner as in the preceding proposition, that this is also the centre of the required circle. N F B A M M B K H G A N E D L F M * This is casily proved indirectly, by means of the seventh proposition of the third book. PROP. VIII. Prob.—To describe a circle touching three given circles.. Let AB, CD, EF be three given circles: it is required to describe a circle touching them. From the centre of CD, with a radius equal to the difference or sum of the radii of AB and CD, describe a circle ; and from the centre of EF, with a radius equal to the difference or sum of the radii of AB and EF, describe a circle: then (APP. II. 5) find the centre of a circle touching the circles now described, and passing through the centre of AB: it will also be the centre of the required circle. The proof is obvious. A B E F B G с E E F Prop. IX. PROB.—Through a given point A, to describe a circle, touching one given straight line BC, and having its centre in another DC. If the lines be parallel, the radius will evidently be their perpendicular distanee asunder, and the construction is obvious. But, if they be not parallel, let them meet in C, and draw a straight line through C and A; through A draw BD perpendicu. lar to BC; from D as centre, at the distance DB describe a circle cutting CA in E ; join ED; and through A draw AF parallel to ED meeting CD in F: F is the centre of the required circle. Draw FG. perpendicnlar to BC. Then, because the triangles CDE, CFA, and CDB, CFG are similar, ED : AF' :: DC : FC, and BD : GF :: DC: FC; whence ED: AF :: BD : GF; and (V. 14) AF, GF are equal, because ED is equal to BD. A circle therefore described from F_as centre, with FG as radius, passes through A; and it touches BC, because the angles at G are right angles. Schol. It is plain, that with the relations of the data in the diagram here employed, there may be two solutions, as either of the points E may be used. Should the circle described with DB as radius, merely touch AC, there would be but one solution: should it not meet CA, the solution would be impossible. PROP. X. THEOR.-If a chord of a given circle have one extremity given in position, and if a segment terminated at that extremity, be taken on the chord, produced if necessary, such that the rectangle under the segment and chord may be equal to a given space; the locus of the point of section is a straight line given in position. |