From any point, as a (Fig. 366), in the tangent bc, draw a line to the centre d; bisect ad at e; upon e, with the radius ea, describe the arc afd; f is the point of contact required.
If f and d were joined, the line would form right angles with the tangent be. (See Art. 352.)
Let a, b and c (Fig. 367) be the three given points. Upon a and b, with any radius greater than half a b, describe arcs intersecting at d and e; upon b and c, with any radius greater than half bc, describe arcs intersecting at f and g; through d and e draw a right line, also another through f and g; upon the intersection h, with the radius ha, describe the circle a bc, and it will be the one required.
Let a b c (Fig. 368) be the given points. Connect them with right lines, forming the triangle acb; bisect the angle cba (Art. 506) with the line bd; also bisect ca in e, and erect ed perpendicular to ac, cutting bd in d; then d is the fourth point required.
A fifth point may be found, as at f, by assuming a, d and b, as the three given points, and proceeding as before. So, also, any number of points may be found simply by using any three already found. This problem will be serviceable in obtaining short pieces of very flat sweeps. (See Art. 240.) The proof of the correctness of this method is found in the fact that equal chords subtend equal angles (Art. 357). Join d and c; then since ae and ec are, by construction, equal, therefore the chords a d and dc are equal; hence the angles they subtend, dba and db c, are equal. So, likewise, chords drawn from a to f, and from f to d, are equal, and subtend the equal angles dbf and f b a. Additional points beyond a or b may be obtained on the same principle. To obtain a point beyond a, on b, as a centre, describe with any radius the arc ion; make on equal to o i; through b and n draw bg; on a as centre and with af for radius, describe the arc, cutting gb at g, then g is the point sought.