PROPOSITION XII. THEOREM. If two circles touch each other externally, the straight line which joins their centres shall pass through the point of contact. Given the two circles ABC, ADE touching each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of ADE; to prove that the straight line which joins the points F and G shall pass through the point of contact A. (Const.) For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG; (Dem.) 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 and AG are equal to FC and DG; wherefore the whole FG is greater than FA and AG; but it is also less (I. 20), which is impossible; A CAD F G E therefore the straight line which joins the centres cannot pass otherwise than through the point of contact A; that is, it must pass through it. One circle cannot touch another in more points than one, whether it touches it internally or externally. join BD, and draw GH bisecting HA B (Const.) For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first internally, in the points B and D ; BD at right angles (I. 10, 11); (Dem.) therefore, because the points B and D are in the circumference of each of the circles, the straight line BD falls within each of them (III. 2); and their centres are in the straight line GH which bisects E B DG A E H BD at right angles (III. 1, Cor.); therefore GH, being the line joining their centres, passes through the point of contact (III. 11); but it does not pass through it, because the points B and D are without the straight line GH, which is absurd; therefore one circle cannot touch another on the inside in more points than one. Nor can two circles touch one another externally in more than for, if it be possible, let the circle ACK touch one point; K the circle ACK is without the circle ABC; and therefore the straight line AC is without this last circle; but because the points A and C are in the circumference of the circle ABC, the straight line AC must be within the same circle, which is absurd; therefore one circle cannot touch another on the outside in more than one point; and it has been shewn that they cannot touch on the inside in more points than one. B Equal chords in a circle are equally distant from the centre; and those which are equally distant from the centre are equal to one another. Given the chords AB and CD, in the circle ABDC, equal to one another; to prove that they are equally distant from the centre. (Const.) Take E the centre of the circle ABDC, and AB double of AF. For A E and from it draw EF and EG, perpendiculars to AB and CD; (Dem.) then, because the straight line EF, passing through the centre, cuts the chord AB, which does not pass through the centre, at right angles, it also bisects it (III. 3); wherefore AF is equal to FB, the same reason, CD is double of CG; and AB is equal to CD; therefore AF is equal to CG. And because AE is equal to EC, the square on AE is equal to the square on EC. But the squares on AF and FE are equal to the square on AE (I. 47), angle AFE is a right angle; and, for a like reason, the squares on EG and GC are equal to the square on EC; therefore the squares on AF and FE are equal to the squares on CG and GE, B D because the of which the square on AF is equal to the square on CG, because AF is equal to CG; therefore the remaining square on FE is equal to the remaining square on EG, and the straight line EF is therefore equal to EG; therefore (III. Def. 15) AB and CD are equally distant from the centre. Next, let it be given that FE is equal to EG; to prove that AB is equal to CD. (Dem.) 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 on EF and FA are equal to the squares on EG and GC; of which the square on FE is equal to the square on EG, because FE is equal to EG; therefore the remaining square on AF is equal to the remaining square on CG; and the straight line AF is therefore equal to CG. And AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Schol. A principle employed in this and the next proposition is-if two quantities A and B be together equal to other two C and D together, that is, if A + B = C + D; then, if A = C, B=D; or, if AC, BD. The diameter is the greatest chord 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. Given a circle ABCD, of which the diameter is AD, and the centre E; and BC a chord nearer to the centre than FG; to prove that AD is greater than any chord BC which is not a diameter, and BC, which is nearer the centre, is greater than FG, which is more remote. (Const.) From the centre draw EH and EK, perpendiculars to BC and FG, and join EB, EC, and EF; because AE is equal to EB, and ED to EC, AD is equal to EB and EC. (Dem.) and and EC are greater than BC (I. 20); fore, also, AD is greater than BC. H K DC And because BC is nearer to the centre than FG, EH is less than EK (III. Def. 15). But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares on EH and HB are equal to the squares on EK and KF, of which the square on EH is less than the square on EK, because EH is less than EK; therefore the square on BH is greater than the square on FK, and the straight line BH greater than FK; and therefore BC is greater than FG. Next, let it be given that BC is greater than FG; to prove that BC is nearer to the centre than FG; that is, the same construction being made, EH is less than EK; (Dem.) because BC is greater than FG, BH likewise is greater than KF; the squares on BH and HE are equal to the squares on FK and KE, of which the square on BH is greater than the square on FK, because BH is greater than FK; therefore the square and on EH is less than the square on EK, EH is less than EK. hence the straight line PROPOSITION XVI. THEOREM. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; (2) and a straight line making an acute angle with the diameter at its extremity cuts the circle. Given a circle ABC, the centre of which is D, and the diameter AB; and AE a line drawn from A per pendicular to AB; to prove that shall fall without the circle. AE join (Const.) In AE take any point F; DF, and let DF meet the circle in C. (Dem.) Because DAF is a right angle, it is greater than the angle AFD (I. 32); therefore DF is greater than DA (I. 19); and DA is equal to DC; therefore DF is greater than DC, and the point F is therefore without the circle. Now, F is any point whatever in the line AE; without the circle. 2. (Const.) Again, let AG be drawn, making the angle DAG less than a right angle; from D draw DH at right angles to AG; (Dem.) and because the angle DHA is a right angle, each of the other angles of the triangle DAH is less than a right angle; the angle DAH is therefore less than the angle DHA, and therefore also the side DH is less than the side DA. Hence the point H is within the circle, and therefore the straight line AG cuts the circle. COR.-From this it is manifest that the straight line which is drawn at right angles to the radius of a circle from the extremity of it, is a tangent; and that it touches it only in one point, because if it did meet the circle in two, it would fall within it (III. 2). Also, it is evident that there can be but one straight line which touches the circle in the same point. To draw a straight line from a given point, either without or in the circumference, which shall be a tangent to a given circle. First, Given the circle BCD and a point A without it; it is required to draw a straight line from A which shall be a tangent to the circle. (Const.) Find the centre E of the circle (III. 1), and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw DF at right angles to EA (I. 11), and join EF and AB. AB touches the circle BCD. therefore the two A (Dem.) Because E is the centre of the circles BCD and AFG; EA is equal to EF, and ED to EB; sides AE and EB are equal to the two FE and ED, and they contain the angle at E common to the two triangles AEB and FED; therefore the base DF is equal to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles (I. 4); fore the angle EBA is equal to the angle EDF. But EDF is a right angle, wherefore EBA is a right angle; and there E BF EB is a radius; but a straight line drawn from the extremity of a radius, at right angles to it, is a tangent (III. 16, Cor.); therefore AB touches the circle, and it is drawn from the given point A, which was required to be done. Second, If the as the point D, angles to DE; given point be in the circumference of the circle, draw the radius DE, and draw DF at right DF touches the circle (III. 16, Cor.). If a straight line be a tangent to a circle, the straight line drawn from the centre to the point of contact shall be perpendicular to the tangent. Given the straight line DE a tangent to the circle ABC at the point C; (Const.) take the centre F, and draw the straight line FC; it is required to prove that FC is perpendi cular to DE. (Dem.) For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a right angle, GCF is an acute angle (I. 17); therefore (I. 19) FC is greater than FG; but FC is equal to FB; therefore FB D A In the same is greater than FG, and it is also less, which is impossible; wherefore FG is not perpendicular to DE. manner it may be shewn, that no other is perpendicular to it besides FC; that is, FC is perpendicular to DE. |