112°. Theorem.-Two circles can intersect in only two points. Proof. If they can intersect in three points, two circles can be made to pass through the same three points. But this is not true. (96°) .. two circles can intersect in only two points. Cor. Two circles can touch in only one point. For a point of contact is equivalent to two points of intersection. 113°. Theorem.-The common centre-line of two intersecting circles is the right bisector Then Cor. 1. By the principle of continuity, OO' always bisects. AB. Let the circles separate until A and B coincide. the circles touch and OO' passes through the point of contact. Def.-Two circles which touch one another have external contact when each circle lies without the other, and internal contact when one circle lies within the other. Cor. 2. Since OO' (Cor. 1) passes through the point of contact when the circles touch one another (a) When the distance between the centres of two circles is the sum of their radii, the circles have external contact. (b) When the distance between the centres is the difference of the radii, the circles have internal contact. (c) When the distance between the centres is greater than the sum of the radii, the circles exclude each other without contact. (d) When the distance between the centres is less than the difference of the radii, the greater circle includes the smaller without contact. (e) When the distance between the centres is less than the sum of the radii and greater than their difference, the circles intersect. I 114°. Theorem.-From any point without a circle two tangents can be drawn to the circle. A S Cor. 1. Since PO is the right bisector of AB, PA=PB. (106°, Cor. 4) (110°, Cor. 3) (113°) (53°) Hence calling the segment PA the tangent from P to the circle, when length is under consideration, we have--The two tangents from any point to a circle are equal to one another. Def. The line AB, which passes through the points of contact of tangents from P, is called the chord of contact for the point P. 115°. Def. 1.—The angle at which two circles intersect is the angle between their tangents at the point of intersection. Def. 2. When two circles intersect at right angles they are said to cut each other orthogonally, The same term is conveniently applied to the intersection of any two figures at right angles. Cor. 1. If, in the Fig. to 114°, PA be made the radius of a circle and P its centre, the circle will cut the circle S orthogonally. For the tangents at A are respectively perpendicular to the radii. Hence a circle S is cut orthogonally by any circle having its centre at a point without S and its radius the tangent from the point to the circle S. 116°. The following examples furnish theorems of some importance. Ex. 1. Three tangents touch the circle S at the points A, B, and C, and intersect to form the ▲A'B'C'. O being the centre of the circle, LAOC=2LA'OC'. A B B A C AC'=BC', and Similarly BA'=CA', (114°, Cor. 1) AAOC'ABOC', and ABOA'=ACOA' LAOC=2LA'OC'. q.e.d. LAOB=24A'OB', and ≤BOC=24B'OC'. If the tangents at A and C are fixed, and the tangent at B is variable, we have the following theorem : The segment of a variable tangent intercepted by two fixed tangents, all to the same circle, subtends a fixed angle at the centre. Ex. 2. If four circles touch two and two externally, the points of contact are concyclic. Let A, B, C, D be the centres of the circles, and P, Q, R, S be the points of contact. Then AB passes through P, BC through Q, etc. (113°, Cor. 1) Now, ABCD being a quadrangle, LA+LB+LC+≤D=4 ̄|s. (90°) But the sum of all the internal angles of the four As APS, BQP, CRQ, and DSR is 8s, and LAPS=LASP, ¿BPQ=¿BQP, etc., LAPS+4BPQ+4CRQ+<DRS = 2 s. SPQ=25-(LAPS+2BPQ), <QRS=2s-(<CRQ+≤DRS), <SPQ+4QRS = 2s. and P, Q, R, S are concyclic (107°). q.e.d. Ex. 3. If the common chord of two intersecting circles subtends equal angles at the two circles, the circles are equal. AB is the common chord, C, C' points upon the circles, and LACB=LAC'B. Let O, O' be the centres. Then LAOB=LAO'B. (106°) And the triangles OAB and O'AB being isosceles are congruent, ... OA=O'A, and the circles are equal. (93, 4) A Ex. 4. If O be the orthocentre of a ▲ABC, the circumcircles to the As ABC, AOB, BOC, COA are all equal. N B S X S2 C ·.· AX and CZ are respectively to BC and AB, LCBA sup. of ≤XOZ = =sup. of ¿COA. But D being any point on the arc AS¿C, LCDA is the sup. of COA. LCBA=LCDA, and the Os S and S, are equal by Ex. 3. In like manner it may be proved that the OS is equal to the Os S3 and S1. Ex. 5. If any point O be joined to the vertices of a ABC, the circles having OA, OB, and OC as diameters intersect upon the triangle. Proof.-Draw OX L to BC and OY L to AC. 40XB, the on OB as diameter passes through X. (107°, Cor. 1) Similarly the ○ on OC as diameter passes through X. Therefore the Os on OB and OC intersect in X; and in like manner it is seen that the Os on OC and OA intersect in Y, and those on OA and OB intersect in Z, the foot of the 1 from O to AB. Ex. 6. The feet of the medians and the feet of the altitudes in any triangle are six concyclic points, and the circle bisects that part of each altitude lying between the orthocentre and the B K (106°, Cor. 1) .. AAFH is isosceles, and AF=FH=FB; LAHB=| (106°, Cor. 4) and H is the foot of the altitude from B. Similarly, K and G are feet of the altitudes from C and A. Again, concyclic (107°), and AO is a diameter of the circumcircle, therefore P is the middle point of AO. Similarly, Q is the middle point of BO, and R of CO. Def.-The circle S passing through the nine points D, E, F, G, H, K, and P, Q, R, is called the nine-points circle of the ДABC. Cor. Since the nine-points circle of ABC is the circumcircle of ▲DEF, whereof the sides are respectively equal to half the sides of the ▲ABC, therefore the radius of the ninepoints circle of any triangle is one-half that of its circumcircle. |