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 produced 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 (II. 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.

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.

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 Cand G touching that circle, straight lines


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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 perpendicular to BF, draw LFAH, through A, 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 Ví. 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. 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.

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, unlessBC, 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.




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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

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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: thereforé 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 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. *

* 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 th

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.)

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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 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 (VI. 16.) the rectangle BL.LG is equal to CL.LF. But (const. and VI. 16.) BL.LG is equal to AL.LM: therefore the rectangles CL.LF, 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.

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 difficulty,

This is easily proved indirectly, by means of the seventh proposition of the third book,




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.







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.


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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.

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