110°. Theorem.-A centre-line and a tangent to the same point on a circle are perpendicular to one another. L' is a centre-line and Ta tangent, both to the point P. is to T. Then L' T B P Cor. 1. Tangents at the end-points of a diameter are parallel. Cor. 2. The perpendicular to a tangent at the point of contact is a centre-line. (Converse of the theorem.) Cor. 3. The perpendicular to a diameter at its end-point is a tangent. D Then and Also and S P Proof.-. T has only the one point P in common with the O, every point of T except P lies without the O... if O is the centre on the line L', OP is the shortest segment from O to T. OP, or L', is to T. (63°, 1) q.e.d. Τ III. Theorem.-The angles between a tangent and a chord from the point of contact are respectively equal to the angles in the opposite arcs into which the chord divides the circle. TP is a tangent and PQ a chord to the same point P, and A is any point on T' the O. Then LQPT=4QAP. Proof. Let PD be a diameter. LQAP LQDP, LDPQ is comp. of QPT, Similarly, the QPT'=LQBP. (106°, Cor. 1) (106°, Cor. 4) (40°, Def. 3) (77°, 6) q.e.d. 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 of their common chord. S O is on the right bisector of AB. B S (54°) Cor. 1. By the principle of continuity, OO' always bisects AB. Let the circles separate until A an B coincide. Then 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. 114°. Theorem.—From any point without a circle two tangents can be drawn to the circle. S A B S Proof. Let S be the and P P the point. Upon the segment PO as diameter let the OS' be described, cutting OS in A and B. Then PA and PB are both tangents to S. For LOAP is in a semicircle and is a .. AP is tangent to S. Similarly BP is tangent to S. (106°, Cor. 4) (110°, Cor. 3) Cor. 1. Since PO is the right bisector of AB, PA=PB. (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, Proof.-and LAOC=2LA'OC'. AC'=BC', BA' CA', (114°, Cor. 1) AAOC'ABOC', and BOA'= COA' LAOB=24A'OB′, and ¿BOC=24B'OC'. S g.e.d. Similarly 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+2B+2C+<D=4 ̄ ̄|s. (90°) But the sum of all the internal angles of the four ▲s APS, BQP, CRQ, and DSR is 8s, and LAPS LASP, LBPQ=4BQP, etc., Now LAPS+4BPQ+<CRQ+4DRS=2s. LSPQ+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. N <SPQ=2s-(<APS+<BPQ), 4QRS=2s-(<CRQ+<DRS), 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) Ex. 4. If O be the orthocentre of a ABC, the circumcircles to the As ABC, AOB, BOC, COA are all equal. B . AX and CZ are respectively to BC and AB, LCBA sup. of LXOZ sup. of COA. X S = But D being any point on the arc AS,C, LCDA is the sup. of COA. LCBA=LCDA, .. S2 and the Os S and S2 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 1 to BC and OY L to AC. |