PROPOSITION 37. THEOREM. If from a point without a circle there are drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle and the part of it without the circle is equal to the square on the line which meets the circle, then the line which meets the circle shall be a tangent to it. E Let ABC be a circle ; and from D, a point without it, let there be drawn two st. lines DCA and DB, of which DCA cuts the circle at C and A, and DB meets it; and let the rect. DA, DC= the sq. on DB. Then shall DB be a tangent to the circle. Construction. From D draw DE to touch the O ABC: III. 17. let E be the point of contact. III. 1. Proof. Since DCA is a secant, and DE a tangent to the circle, the rect. DA, DC = the sq. on DE, III. 36. But, by hypothesis, the rect. DA, DC=the sq. on DB; .. the sq. on DE= the sq. on DB; ... DE = DB. Proved. 1. Def. 15. and DF is common ; - DBF=the _ DEF. But DEF is a rt. angle, for DE is a tangent; III. 18. .:. DBF is also a rt. angle; and since BF is a radius, .. DB touches the O ABC at the point B. Q.E.D. .. the I. 8. NOTE ON THE METHOD OF LIMITS AS APPLIED TO TANGENCY. Euclid defines a tangent to a circle as a straight line which meets the circumference, but being produced, does not cut it: and from this definition he deduces the fundamental theorem that a tangent is perpendicular to the radius drawn to the point of contact. III. Prop. 16. But this result may also be established by the Method of Limits, which regards the tangent as the ultimate position of a secant when its two points of intersection with the circumference are brought into coincidence [See Note on page 165] : and it may be shewn that every theorem relating to the tangent may be derived from some more general proposition relating to the secant, by considering the ultimate case when the two points of intersection coincide. 1. To prove by the Method of Limits that a tangent to a circle is at right angles to the radius drawn to the point of contact. Let ABD be a circle, whose centre is C; and PABQ a secant cutting the Oce in A and B; and let P'AQ' be the limiting position of PQ when the point B is brought into coincidence with A. Then shall CA be perp. to P'Q'. Bisect AB at E and join CE: then CE is perp. to PQ. III. 3. Now let the secant PABQ change its position in such a way that while the point A remains fixed, the point B continually approaches A, and ultimately coincides with it; then, however near B approaches to A, the st. line CE is always perp. to PQ, since it joins the centre to the middle point of the chord AB. But in the limiting position, when B coincides with A, and the secant PQ becomes the tangent P'Q', it is clear that the point E will also coincide with A; and the perpendicular CE becomes the radius CA. Hence CA is perp. to the tangent P'Q' at its point of contact A. Q.E.D. NOTE. It follows from Proposition 2 that a straight line cannot cut the circumference of a circle at more than two points. Now when the two points in which a secant cuts a circle move towards coincidence, the secant ultimately becomes a tangent to the circle : we infer therefore that a tangent cannot meet a circle otherwise than at its point of contact. Thus Euclid's definition of a tangent may be deduced from that given by the Method of Limits. ए P III. 21. 2. By this method Proposition 32 may be derived as a special case from Proposition 21. For let A and B be two points on the Oce of the O ABC; and let BCA, BPA be any two angles in the segment BCPA: then the L BPA=the L BCA. Produce PA to Q. Now let the point P continually approach the fixed point A, and ultimately coincide with it; A then, however near P may approach to A, the L BPQ=the LBCA. III. 21. and the secant PAQ becomes the tangent AQ', and the L BPQ becomes the L BAQ'. Hence the L BAQ'=the L BCA, in the alternate segment. Q.E.D. The contact of circles may be treated in a similar manner by adopting the following definition. DEFINITION. If one or other of two intersecting circles alters its position in such a way that the two points of intersection continually approach one another, and ultimately coincide; in the limiting position they are said to touch one another, and the point in which the two points of intersection ultimately coincide is called the point of contact. EXAMPLES ON LIMITS. 1. Deduce Proposition 19 from the Corollary of Proposition 1 and Proposition 3. 2. Deduce Propositions 11 and 12 from Ex. 1, page 171. 5. Shew that a straight line cuts a circle in two different points, two coincident points, or not at all, according as its distance from the centre is less than, equal to, or greater than a radius. 6. Deduce Proposition 32 from Ex. 3, page 202. 7. Deduce Proposition 36 from Ex. 7, page 227. 8. The angle in a semi-circle is a right angle. To what Theorem is this statement reduced, when the vertex of the right angle is brought into coincidence with an extremity of the diameter ? 9. From Ex. 1, page 204, deduce the corresponding property of a triangle inscribed in a circle. THEOREMS AND EXAMPLES ON BOOK III. I. ON THE CENTRE AND CHORDS OF A CIRCLE. [See Propositions 1, 3, 14, 15, 25.] 1. All circles which pass through a fixed point, and have their centres on a given straight line, pass also through a second fixed point. Let AB be the given st. line, and P the given point. B R Р A From P draw PR perp. to AB; Join CP, CP. CR is common, Constr. 1. 4. :: the circle whose centre is C, and which passes through P, must pass also through P'. But C is the centre of any circle of the system; :. all circles, which pass through P, and have their centres in AB, pass also through p?. Q. E. D. 2. Describe a circle that shall pass through three given points not in the same straight line. 3. Describe a circle that shall pass through two given points and have its centre in a given straight line. When is this impossible ? 4. Describe a circle of given radius to pass through two given points. When is this impossible? 5. ABC is an isosceles triangle; and from the vertex A, as centre, a circle is described cutting the base, or the base produced, at X and Y. Shew that BX=CY. 6. If two circles which intersect are cut by a straight line parallel to the common chord, shew that the parts of it intercepted between the circumferences are equal. 7. If two circles cut one another, any two straight lines drawn through a point of section, making equal angles with the common chord, and terminated by the circumferences, are equal. (Ex. 12, p. 171.] 8. If two circles cut one another, of all straight lines drawn through a point of section and terminated by the circumferences, the greatest is that which is parallel to the line joining the centres. 9. Two circles, whose centres are C and D, intersect at A, B ; and through A a straight line PAQ is drawn terminated by the circumferences: if PC, QD intersect at X, shew that the angle PXQ is equal to the angle CAD. 10. Through a point of section of two circles which cut one another draw a straight line terminated by the circumferences and bisected at the point of section. 11. AB is a fixed diameter of a circle, whose centre is C; and from P, any point on the circumference, PQ is drawn perpendicular to AB; shew that the bisector of the angle CPQ always intersects the circle in one or other of two fixed points. 12. Circles are described on the sides of a quadrilateral as diameters: shew that the common chord of any two consecutive circles is parallel to the common chord of the other two. [Ex. 9, p. 105.] 13. Two equal circles touch one another externally, and through the point of contact two chords are drawn, one in each circle, at right angles to each other : shew that the straight line joining their other extremities is equal to the diameter of either circle. 14. Straight lines are drawn from a given external point to the circumference of a circle: find the locus of their middle points. [Ex. 3, p. 105.] 15. Two equal segments of circles are described on opposite sides of the same chord AB; and through O, the middle point of AB, any straight line POQ is drawn, intersecting the arcs of the segments at P and Q: shew that OP=OQ. |