COR. 2. And, conversely, if the distance between the centres be equal to the difference of the radii, the two circles will touch each other internally. PROP. XII. THEOR. If two circles touch each other externally, the straight line which joins their centres will pass through the point of contact. Let the two circles ABC, ADE, touch each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of ADE; The straight line which joins the points F, G, shall pass through the point of contact. B E For, if not, let it pass otherwise, if possible, FCDG, and join FA, AG: and because F is the centre of the circle ABC, AF is equal to FC: Also because G is the centre of the circle ADE, AG is equal to GD. Therefore FA, AG are equal to FC, DG; wherefore the whole FG is greater than FA, AG; but it is also less (13. 1.), which is impossible: Therefore the straight line which joins the points F, G cannot pass otherwise than CAD F through the point of contact A; that is, it passes through A. COR. Hence, if two circles touch each other externally, the distance between their centres will be equal to the sum of their radii. And, conversely, if the distance between the centres be equal to the sum of the radii, the two circles will touch each other externally. SCHOLIUM. In order to express the Scholium and Corollaries to Propositions X., XI., XII., algebraically; let R and r denote the radii of the two circles, and d the distance between their centres. 1. If the two circles cut; then, d<R+r. And, conversely, if d<R+r, the two circles will cut each other. In both cases, R<d+r; where R denotes the radius of the greater circle. 2. If the two circles will touch each other internally, then, d=R—r. And, conversely, if d=R―r, the two circles will touch each other internally. 3. If two circles touch each other externally; then, d=R+r. And, conversely, if d=R+r, the two circles will touch each other externally. 4. If d>R+r, or d<R―r, the circumferences of the two circles will neither cut nor touch each other. PROP. XIII. THEOR. In the same circle, equal angles at the centre are subtended by equal arcs ; and, conversely, equal arcs subtend equal angles at the centre. Let C be the centre of a circle, and let the angle ACD be equal to the angle BCD; then the arcs AFD, DGB, subtending these angles, are equal. Join AD, DB; then the triangles ACD, BCD, having two sides and the included angle in the one, equal to two sides and the included angle in the other, are equal: so that, if ACD be applied to BCD, there shall be an entire coincidence, the point A coinciding with B, and D common to both arcs; the two extremities, therefore, of the arc AFD, thus coinciding with those of the arc BGD, all the intermediate parts must coincide, inasmuch as they are all equally distant from the centre. A H E B FD G Conversely. Let the arc AFD be equal to the arc BGD; then the angle ACD is equal to the angle BCD. For, if the arc AFD be applied to the arc BGD, they would coincide; so that the extremities AD of the chord AD, would coincide with those of the chord BD; these chords are therefore equal: hence, the angle ACD is equal to the angle BCD (Th. V. B. I.). COR. 1. It follows, moreover, that equal angles at the centre are subtended by equal chords: and, conversely, equal chords subtend equal angles at the centre. COR. 2. It is also evident, that equal chords subtend equal arcs; and conversely, equal arcs are subtended by equal chords. COR. 3. If the angle at the centre of a circle be bisected, both the arc and the chord which it subtends shall also be bisected. COR. 4. It follows, likewise, that a perpendicular through the middle of the chord, bisects the angle at the centre, and passes through the middle of the arc subtended by that chord. SCHOLIUM. The centre C, the middle point E of the chord AB, and the middle point D of the arc subtended by this chord, are three points situated in the same line perpendicular to the chord. But two points are sufficient to determine the position of a straight line; hence every straight line which passes through two of the points just mentioned, will necessarily pass through the third, and be perpendicular to the chord. PROP. XIV. THEOR. Equal straight lines in a circle are equally distant from the centre; and those which are equally distant from the centre, are equal to one another. Let the straight lines AB, CD, in the circle ABDC, be equal to one another they are equally distant from the centre. : A E Take E the centre of the circle ABDC, and from it draw EF, EG, perpendiculars to AB, CD; join AE and EC. Then, because the straight line EF passing through the centre, cuts the straight line AB, which does not pass through the centre at right angles, it also bisects (3. 3.) it: Wherefore AF is equal to FB, and AB double of AF. For the same reason, CD is double of CG: But AB is equal to CD; therefore AF is equal to CG: And because AE is equal to EC, the square of AE is equal to the square of EC: Now the squares of AF, FE are equal (37. 1.) to the square of AE, because the angle AFE is a right angle; and, for the like reason, the squares of EG, GC, are equal to the square of EC: Therefore the squares of AF, FE are equal to the squares of CG, GE, of which the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of FE is equal to the remaining square of EG, and the straight line EF is therefore equal to EG: But straight lines in a circle are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal (4. Def. 3.): therefore AB, CD are equally distant from the centre. B D Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG, AB is equal to CD. For, the same construction being made, it may, as before, be demonstrated, that AB is double of AF, and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GC; of which the square of FE is equal to the square of EG, because FE is equal to EG; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG: But AB is double of AF, and CD double of CG; wherefore AB is equal to CD. PROP. XV. THEOR. The diameter is the greatest straight line in a circle; and of all others, that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the less. Let ABCD be a circle, of which the diameter is AD, and the centre E; and let BC be nearer to the centre than FG; AD is greater than any straight line BC which is not a diameter, and BC greater than FG. From the centre draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF; and because AE is equal to EB, and ED to EC, AD is equal to EB, EC: But EB, EC are greater (13. 1.) than BC; wherefore, also, AD is greater than BC. And, because BC is nearer to the centre than F FG, EH is less (5. Def. 3.) than EK; But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares of EK, KF, of which the square of EH is less than the square of EK, because EH is less than EK; therefore the square of BH is greater than the K square of FK, and the straight line BH greater than FK; and therefore BC is greater than FG. Next, let BC be greater than FG; BC is nearer to the centre than FG: that is, the same construction being made, EH is less than EK; Because BC is greater than FG, BH likewise is greater than KF but the squares of BH, HE are equal to the squares of FK, KE, of which the square of BH is greater than the square of FK, because BH is greater than FK; therefore the square of EH is less than the square of EK, and the straight line EH less than EK. COR. The shorter the chord is, the farther it is from the centre; and, conversely, the farther the chord is from the centre, the shorter it is. PROP. XVI. THEOR. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn between that straight line and the circumference, from the extremity of the diameter, so as not to cut the circle. Let ABC be a circle, the centre of which is D, and the diameter AB : and let AE be drawn from A perpendicular to AB, AE shall fall without the circle. In AE take any point F, join DF and let DF meet the circle in C. Because DAF is a right angle, it is greater than the angle AFD (3 Cor. Th. 25. 1.); but the greater angle of any triangle is subtended by the greater side (12. 1.), therefore DF is greater than DA; now DA is equal to DC, therefore DF is greater than DC, and the point F is therefore without the circle. And F is any point whatever in the line AE, therefore AE falls without the circle. Again, between the straight line AE and the circumference, no straight line can be drawn from the point A, which does not cut the circle. Let AG be B D A of it, touches the circle; and that it touches it only in one point; because, if it did meet the circle in two, it would fall within it (2. 3.). Also it is evident that there can be but one straight line which touches the circle in the same point. COR. 2. Hence, a perpendicular at the extremity of a diameter is a tangent to the circle; and, conversely, a tangent to a circle is perpendicular to the diameter drawn from the point of contact. COR. 3. It follows, likewise, that tangents at each extremity of the diameter are parallel (Cor. Th. 20. B. I.); and, conversely, parallel tangents are both perpendicular to the same diameter, and have their points of contact at its extremities. PROP. XVII. THEOR. The arcs of a circle intercepted by two parallels are equal; and, conversely, if two straight lines intercept equal arcs of a circle, and do not cut each other within the circle, the lines will be parallel. There may be three cases First. If the parallels are tangents to the circle, as AB, CD; then, each A Second. When, of the two parallels AB, GH, one is a tangent, the other a chord, which being perpendicular to FE, the arc GEH is bisected by FE (Th. 13. Cor. 4. B. 3.); so that in this case also, the intercepted arcs GE, EH are equal. Third. If the two parallels are chords, as GH, JK; let the diameter FE be perpendicular to the chord GH, it will also be perpendicular to JK, since they are parallel; therefore, this diameter must bisect each of the arcs which they subtend: that is, GE=EH, and JE=EK; therefore, JE-GE=EK-EH; or, which amounts to the same thing, JG is equal to HK. |