From O draw a tangent to this circle, touching it in P. Join O'P, and let it, produced through P if necessary, meet the circumference of the circle whose centre is O' in the point Q. Through O draw OR parallel to PQ, and join QR. QR will be a tangent to both circles. Proof. Since PQ is by the construction equal and parallel to OR, therefore RQ is parallel to OP. But OP is at right angles to O'P, since it touches the circle in P, and therefore RQ is at right angles to OR and OQ; and therefore touches both circles. COR. I. When the circles are wholly outside one another, they have four common tangents: when they touch externally, they have three common tangents: when they intersect one another, they have two common tangents: when they touch internally, they have one common tangent: and when one of the circles is wholly inside the other, they have no common tangent. COR. 2. By symmetry with respect to 00', pairs of the common tangents will intersect on that line. PROBLEM 6. To find the locus of the centres of circles which touch two given straight lines. Let AOB, COD be the two given straight lines, intersecting one another in O. Let P be the centre of a circle which touches both the lines, PN, PM the perpendiculars from P on DOC and АОВ. Then PN=PM, and if OP be joined, since the triangles ONP, OMP are right angled at N and M, have the A M B D hypothenuse OP common, and have one side PN= one side PM, being radii of the same circle; therefore the triangles are equal in all respects, and the angle PON = the angle POM, that is OP bisects the angle COB. Therefore the centres lie on the bisectors of the angles between the given lines; and these bisectors are therefore the locus required. COR. I. It is obvious that these bisectors form two straight lines at right angles to one another. COR. 2. If the given lines are parallel, the locus is a line parallel to both and equidistant from them. PROBLEM 7. To describe a circle to touch three given straight lines of indefinite length. Let the three given lines intersect in A, B, and C. 03 Then since the circle required is to touch the lines that intersect in A, its centre must lie on one of the bisectors of the angles at A. Similarly, it must lie on one of the bisectors of the angles at B. Therefore the construction is suggested. Construction. Draw the bisectors of the angles at A and B, which will intersect in four points O, 01, 02, 03. These will be the centres of the circles required, and a circle described with any one of these points as centre, to touch one of the given lines, will touch the other two. COR. I. It follows that COO, and O,CO, are straight lines, that is, the six bisectors of the interior and exterior angles of a triangle intersect one another three and three in four points. COR. 2. If two of the lines are parallel, only two circles can be described to touch the three lines. COR. 3. If all the lines are parallel, no circle can be described to touch them all. SECTION V. REGULAR POLYGONS. One of the most interesting species of problem connected with the circle consists in describing regular polygons, and inscribing them in, or circumscribing them about given circles. We shall first establish the following theorems. THEOREM 12. If from the centre of a circle radii are drawn to make equal angles with one another consecutively all round, then if their extremities are joined consecutively, a regular polygon will be inscribed in the circle, and if at their extremities, tangents are drawn, a regular polygon will be circumscribed to the circle. Let OA, OB, OC, OD... be radii making equal angles consecutively to one another. (1) Join AB, BC, CD... Then since the angles at O are equal, the chords which subtend them are equal, and therefore the inscribed polygon is equilateral. B D R And since the arcs AB, BC, CD ... are equal; therefore the arc AC = arc BD, and therefore the angle ABC in the segment ABC= the angle BCD in the segment BCD; that is, the polygon is equiangular. Hence ABCD... will be a regular polygon inscribed in the circle. (2) Draw tangents at A, B, C, D... meeting one another in P, Q, R... Join PO, QO. Since APPB, the line PO bisects the angle AOB, and similarly QO bisects the angle BOC. Therefore the angle POBthe angle QOB, and hence PB = BQ from the triangles POB, QOB. Therefore QP is double of QB; and similarly QR is double of QC. But QBQC; and therefore QP=QR; and thus it may be shewn that the polygon is equilateral. And since the angles APB, BQC... are supplementary to the angles AOB, BOC... they are equal to one another. That is, the polygon is equiangular; and since PQ, QR... are tangents, PQR... will be a regular polygon circumscribed to the circle. THEOREM 13. In a regular polygon the bisectors of the angles intersect in one point, which is the centre of the circles inscribed in the polygon, or circumscribed about it. Let ABCDE... be a regular polygon. Bisect the angles B, C by straight lines meeting in O. Join OD, OE... Since BO, CO are bisectors of the equal angles ABC, BCD, therefore the angle OBC= the angle OCB; and therefore OB = OC. P R B Q C E And because in the triangles OCB, OCD, the angle |