Page images


Through three given points not in the same straight line, one circumference may always be made to pass, and but one.

[blocks in formation]

For, they must necessarily cut A each other, if they are not parallel.


Now, if they were parallel, the line AB, which is perpendicular to DE, would also be perpendicular to FG, and the angle K would be a right angle (Book I. Prop. XX. Cor. 1.). But BK, the prolongation of BD, is a different line from BF, because the three points A, B, C, are not in the same straight line; hence there would be two perpendiculars, BF, BK, let fall from the same point B, on the same straight line, which is impossible (Book I. Prop. XIV.); hence DE, FG, will always meet in some point O.

And moreover, this point O, since it lies in the perpendicular DE, is equally distant from the two points, A and B (Book 1. Prop. XVI.); and since the same point O lies in the perpendicular FG, it is also equally distant from the two points B and C: hence the three distances OA, OB, OC, are equal; therefore the circumference described from the centre O, with the radius OB, will pass through the three given points A, B, C.

We have now shown that one circumference can always be made to pass through three given points, not in the same straight line: we say farther, that but one can be described through them.

For, if there were a second circumference passing through the three given points A, B, C, its centre could not be out of the line DE, for then it would be unequally distant from A and B (Book I. Prop. XVI.); neither could it be out of the line FG, for a like reason; therefore, it would be in both the lines DE, FG. But two straight lines cannot cut each other in more than one point; hence there is but one circumference which can pass through three given points.

Cor. Two circumferences cannot meet in more than two points; for, if they have three common points, there would be two circumferences passing through the same three points; which has been shown by the proposition to be impossible.


Two equal chords are equally distant from the centre; and of two unequal chords, the less is at the greater distance from the


First. Suppose the chord AB= DE. Bisect these chords by the perpendiculars CF, CG, and draw the radii CA, CD.




In the right angled triangles CAF, DCG, the hypothenuses CA, CD, are equal; and the side AF, the half of AB, is equal to the side DG, the half of DE: hence the triangles are equal, and CF is equal to CG (Book I. Prop. XVII.); hence, the two equal chords AB, DE, are equally distant from the centre.

Secondly Let the chord AH be greater than DE. The arc AKH will be greater than DME (Prop. V.): cut off from the former, a part ANB, equal to DME; draw the chord AB, and let fall CF perpendicular to this chord, and CI perpendicular to AH. It is evident that CF is greater than CO, and CO than CI (Book I. Prop. XV.); therefore, CF is still greater than CI. But CF is equal to CG, because the chords AB, DE, are equal: hence we have CG>CI; hence of two unequal chords, the less is the farther from the centre.


Let BD be perpendicular to the B radius CA, at its extremity A; then will it be tangent to the circumfe




[ocr errors]

A straight line perpendicular to a radius, at its extremity, is a tangent to the circumference.




For, every oblique line CE, is longer than the perpendicular CA (Book I. Prop. XV.); hence the

point E is without the circle; therefore, BD has no point but A common to it and the circumference; consequently BD is a tangent (Def. 8.).

Scholium. At a given point A, only one tangent AD can be drawn to the circumference; for, if another could be drawn, it would not be perpendicular to the radius CA (Book I. Prop. XIV. Sch.); hence in reference to this new tangent, the radius AC would be an oblique line, and the perpendicular let fall from the centre upon this tangent would be shorter than CA; hence this supposed tangent would enter the circle, and be a



Two parallels intercept equal arcs on the circumference.

There may be three cases.

First. If the two parallels are secants, draw the radius CH perpendicular to the chord MP. It will, at the same time be perpendicular to NQ (Book J.Prop.XX.Cor.1.); therefore, the point H will be at once the middle of the arc MHP, and of the arc NHQ (Prop. VI.); therefore, we shall have the arc MH=HP, and the arc NH=

HQ; and therefore MH-NH=HP-HQ; in other words, MN=PQ:


Second. When, of the two parallels AB, DE, one is a secant, the other a tangent, draw the radius CH Ato the point of contact H; it will be perpendicular to the tangent DE (Prop. IX.), and also to its parallel MP. But, since CH is perpendicular to the chord MP, the point H must be the middle of the arc MHP (Prop. VI.); therefore the arcs MH, HP, included between the parallels AB, DE, are equal.















Third. If the two parallels DE, IL, are tangents, the one at H, the other at K, draw the parallel secant AB; and, from what has just been shown, we shall have MH=HP, MK=KP; and hence the whole arc HMK=HPK. It is farther evident that each of these arcs is a semicircumference.


If two circles cut each other in two points, the line which passes through their centres, will be perpendicular to the chord which joins the points of intersection, and will divide it into two equal parts.

For, let the line AB join the points of intersection. It will be a common chord to the two circles. Now if a perpendicular

be erected from the middle of this chord, it will pass through each of the two centres C and D (Prop. VI. Sch.). But no more than one straight line can be drawn through two points; hence the straight line, which passes through the centres, will bisect the chord at right angles.


If the distance between the centres of two circles is less than the sum of the radii, the greater radius being at the same time less than the sum of the smaller and the distance between the centres, the two circumferences will cut each other.

For, to make an intersection possible, the triangle CAD must be possible. Hence, not only must we have CD<AC+AD, but also the greater radius AD< AC+CD (Book I. Prop. VII.). And, whenever the triangle CAD can be constructed, it is plain

that the circles described from the centres C and D, will cut each other in A and B.




If the distance between the centres of two circles is equal to the sum of their radii, the two circles will touch each other externally.

Let C and D be the centres at a distance from each other equal to CA+AD.

The circles will evidently have the point A common, and they will have no other; because, if they had two points common, the distance between their centres must be less than the sum of their radii.



[ocr errors]

If the distance between the centres of two circles is equal to the difference of their radii, the two circles will touch each other · internally.

Let C and D be the centres at a dis- E tance from each other equal to AD-CA.

It is evident, as before, that they will have the point A common; they can have no other; because, if they had, the greater radius AD must be less than the sum of the radius AC and the distance CD between the centres (Prop. XII.); which is contrary to the supposition.

Cor. Hence, if two circles touch each other, either externally or internally, their centres and the point of contact will be in the same right line.

Scholium. All circles which have their centres on the right line AD. and which pass through the point A, are tangent to each other. For, they have only the point A common, and if through the point A, AE be drawn perpendicular to AD, the straight line AE will be a common tangent to all the circles.

« PreviousContinue »