IESSONS IN GEOMETRY. 



LESSONS IN GEOMETRY. XV. 



\ X V. To find the centre of any given circle, or of 



'/e. 



Let ABC (Fig. 53) bo the given circle of which it in T> 

 t> fm<l the oontro. First, draw any straight lino, A B, dividing 

 the circle into two unequal segments. Bisect A B in D, and 

 through the point D draw the straight lino K c at right angles 

 to A B. Bisect E c in F. The point F is the centre of the circle 



ABC. 



There are other methods by which the centre of the circle 

 ABC may bo found, although tho one that has just been described 

 is i .crimps the most simple. For instance, we might .have drawn 

 the straight lines a H, x L as tangents to the circle A B c, through 



tho points A and B, and at 

 the points of contact, A and 

 B, drawn tho straight linos 

 AN, BO, at right angles to 

 the straight linos OH, K i.. 

 and intersecting each other 

 in the point F; from which 

 we learn that if any two 

 points be taken in the cir- 

 cumference of a circle, which 

 are not the opposite extremi- 

 ties of a diameter of that 

 circle, and tangents to the 

 circle be drawn through 

 these points, the straight 

 lines drawn at right angles 

 to the tangents through the 

 points of contact shall inter- 

 sect each other, if produced 

 far enough, in the centre of 

 the circle. 



This method is useful when wo wish to find the centre from 

 which an arc or part of the circumference of a circle of very 

 great extent has been described. The following is a third 

 method of finding the centre of a given circle or the given arc of 

 any circle. Let us suppose, as before, that A B c in Fig. 53 

 represents the given circle. Set off along any part of tho cir- 

 cumf eroace three equal arcs, BE, E A, and A p. Then from tho 

 points P and E as centres, with any radius greater than the 

 radius of the given circle, describe two arcs intersecting each 

 other in the point N ; and from the points A and B as centres, 

 with any radius greater than the radius of tho given circle, 

 describe two arcs intersecting each other in the point Q. Join 

 AN, E Q. The point F in which these lines intersect each other 

 ia the centre of the circle ABC. 



Our figures, as we have said before, sometimes appear compli- 

 cated from the necessity that there is of saving as much space 

 as we can by making one diagram serve as an 

 illustration either to many methods of doing tho 

 same thing, or to sequences that may arise out 

 of the consideration of the problem in question. 

 Our readers are therefore in all cases when it is 

 necessary recommended to study our problems 

 with a piece of paper, a pair of compasses, and 

 a parallel ruler at hand, that they may construct 

 for themselves just so much of our diagram as 

 is necessary for an illustration of the process in 

 course of description, disentangling it as it were from tho 

 figure that we have given as a means of explaining our direc- 

 tions. As an example of this, wo give in Fig. 54, on a reduced 

 scale, just so much ns is absolutely necessary of Fig. 53 to 

 enable a reader to understand the first method that we have 

 given of finding the centre of any given circle. 



Some of tho methods that have been described for finding the 

 centre of a given circle apply equally well, as it may have been 

 seen, to finding the centre from which any given arc of a circle 

 has been described ; but there is another method of finding the 

 centre of any given arc that we will now proceed to bring under 

 the reader's notice. 



First, lot A c B in Fig. 53 be tho arc of which it is required to 

 find tho centre. Join A B ; bisect A B in D ; draw D c at right 

 angles to A B, and join A c. Then at the point A in the straight 

 line c A make the angle c A F equal to the angle A c F, and pro- 

 duce the leg A F of the angle c A F, if necessary, far enough to 

 30 N.K, 



Fig. 54. 



intersect c D in r. Tho point F U the centre from which tin 

 arc A c B has been described. 



Now let A E B in Fig. 53 b i the arc of which it u required to 

 find tho centre. Join A B ax before ; bisect A B in u. Draw 

 D K perpendicular to A B, and join A t. Produce X D indefinitely 

 towards c, and at the point A in the straight line X A, make tho 

 angle x A F equal to the angle A B F, producing tLe leg A r of 

 tho angle E A F, if necessary, far enough to intersect D pro- 

 duced in tho point r. This point, u before, U the centre from 

 which tho arc A E B has been described. 



In the first of these two cases it will be noticed that the arc 

 of which the centre is required is greater than half the circum- 

 ference of the circle of which it is an arc, but in the second it is 

 less than half the circumference. If the arc were half the 

 circumference, it is plain that to find its centre all we hare to 

 do is to join its extremities, and bisect the chord that joins 

 them. 



On further inspection of Fig. 53 it will bo noticed that the 

 straight lines OH, K L, v'uch were drawn as tangents to the 

 circle ABC through tho points A and B, have their points of 

 intersection M in the straight line c B obtained by producing 

 c E in an upward direction ; and the angle A u c is equal to the 

 angle BMC. This leads to another mode of finding the centre 

 of the circle ABC, which is as follows : 



Through any two points, A and B, in the circumference of the 

 given circle ABC, draw tho tangents OH, K L, intersecting each 

 other in the point M. Bisect the angle A M B by the straight 

 line H E, and produce it to cut the circumference of the circle 

 in c. Bisect c E in F. The point F, as before, is the centre of 

 tho circle ABC. 



PROBLEM XXXVI. To describe a circle through any three 

 given points which are not in. the same 

 straight line. 



Let A, B, c (Fig.55), be the three given 

 points through which it is required 

 to describe a circle, or rather the cir- 

 cumference of a circle. Join A B, A c, 

 and bisect these straight lines respec- 

 tively in the points D and E. Through 

 D draw the straight line D F of inde- 

 finite length, perpendicular to A B, 

 and through E draw the straight line 

 E o, also of an indefinite length, per- 

 pendicular to A c. Tho point of in- 

 tersection, H, of the straight lines D F, 

 E o, is the centre from which a circle 

 may be described with a radius, H A, that shall pass through 

 the other two given points, A, B, and c. The same result would 

 be obtained by joining the straight lines A B, B c, or A c, c B, 

 bisecting them, and drawing perpendiculars through the points 

 of bisection as shown in the figure. 



PROBLEM XXXVII. To draw a tangent to a given circlf 

 through any given point either in the circumference of the circle 

 or vitlwut it. 



The case in which the given point is in tho circumference of 



the circle needs no illus 

 tration and very little 

 explanation, for it is 

 manifest that nothing 

 more is required than to 

 draw a straight line join- 

 ing the centre of tho 

 circle and the given point, 

 and then through tho 

 given point to draw a 

 straight line at right 

 angles to the radius ot 

 the circle thus obtained. 

 The straight lino drawn through the given point at right angle* 

 to the radius will be a tangent to the given circle. 



In the case in which the given point lies without the circum- 

 ference of the circle, let A B c (Fig. 56) represent the given circle, 

 and D the given point without it. Find E, the centre of tho 

 circle ABC, and join D E. Bisect D X in F. and from the point 

 F as centre, at the distance F x or r t as radius, describe the 

 circle D o H, cutting the circumference of the circle A B c in tho 

 points o, H. Join D o, D H, and produce them indefinitely 

 towards K and i. respectively. The straight lines D K, D L are 



n* 



Fi(f. 



