1 EXERCISES ON THE CONTACT OF CIRCLES. (Numerical and Graphical.) From centres 2.6′′ apart draw two circles with radii 1.7" and 0.9" respectively. Why and where do these circles touch one another? If circles of the above radii are drawn from centres 0.8" apart, prove that they touch. How and why does the contact differ from that in the former case? 2. Draw a triangle ABC in which a=8 cm., b=7 cm., and c=6 cm. From A, B, and C as centres draw circles of radii 2.5 cm., 3.5 cm., and 4.5 cm. respectively; and shew that these circles touch in pairs. 3. In the triangle ABC, right-angled at C, a=8 cm. and b=6 cm. ; and from centre A with radius 7 cm. a circle is drawn. What must be the radius of a circle drawn from centre B to touch the first circle? 4. A and B are the centres of two fixed circles which touch internally. If P is the centre of any circle which touches the larger circle internally and the smaller externally, prove that AP+BP is constant. If the fixed circles have radii 50 cm. and 30 cm. respectively, verify the general result by taking different positions for P. 5. AB is a line 4" in length, and C is its middle point. On AB, AC, CB semicircles are described. Shew that if a circle is inscribed in the space enclosed by the three semicircles its radius must be ". (Theoretical.) 6. A straight line is drawn through the point of contact of two circles whose centres are A and B, cutting the circumferences at P and Q respectively; shew that the radii AP and BQ are parallel. 7. Two circles touch externally, and through the point of contact a straight line is drawn terminated by the circumferences: shew that the tangents at its extremities are parallel. 8. Find the locus of the centres of all circles (i) which touch a given circle at a given point; (ii) which are of given radius and touch a given circle. 9. From a given point as centre describe a circle to touch a given circle. How many solutions will there be? 10. Describe a circle of radius a to touch a given circle of radius b at a given point. How many solutions will there be? THEOREM 49. [Euclid III. 32.] The angles made by a tangent to a circle with a chord drawn from the point of contact are respectively equal to the angles in the alternate segments of the circle. Let EF touch the O ABC at B, and let BD be a chord drawn from B, the point of contact. It is required to prove that (i) the FBD = the angle in the alternate segment BAD: Let BA be the diameter through B, and C any point in the arc of the segment which does not contain A. Join AD, DC, CB. Proof. Because the ADB in a semi-circle is a rt. angle, .. the FBA the 2 DBA, BAD together. then the FBD = Take away the common DBA, = the BAD, which is in the alternate segment Again because ABCD is a cyclic quadrilateral, .. the EBD = the BCD, which is in the alternate segment. Q.E.D. EXERCISES ON THEOREM 49. 1. In the figure of Theorem 49, if the LFBD=72°, write down the values of the L BAD, BCD, EBD. 2. Use this theorem to shew that tangents to a circle from an external point are equal. X3. Through A, the point of contact of two circles, chords APQ, AXY are drawn: shew that PX and QY are parallel. Prove this (i) for internal, (ii) for external contact. 4. AB is the common chord of two circles, one of which passes through O, the centre of the other: prove that OA bisects the angle between the common chord and the tangent to the first circle at A. 5. Two circles intersect at A and B; and through P, any point on one of them, straight lines PAC, PBD are drawn to cut the other at C and D: shew that CD is parallel to the tangent at P. 6. If from the point of contact of a tangent to a circle a chord is drawn, the perpendiculars dropped on the tangent and chord from the middle point of either arc cut off by the chord are equal. EXERCISES ON THE METHOD OF LIMITS. 1. Prove Theorem 49 by the Method of Limits. [Let ACB be a segment of a circle of which AB is the chord; and let PAT' be any secant through A. Join PB. Then the BCA = the LBPA; Theor. 39. and this is true however near P approaches to A. If P moves up to coincidence with A, then the secant PAT' becomes the tangent AT, and the LBPA becomes the L BAT. .., ultimately, the LBAT = the BCA, in the alt. segment.] 2. From Theorem 31, prove by the Method of Limits that P The straight line drawn perpendicular to the diameter of a circle at its extremity is a tangent. 3. Deduce Theorem 48 from the property that the line of centres bisects a common chord at right angles. 4. Deduce Theorem 49 from Ex. 5, page 163. 5. Deduce Theorem 46 from Theorem 41. PROBLEMS. GEOMETRICAL ANALYSIS. Hitherto the Propositions of this text-book have been arranged Synthetically, that is to say, by building up known results in order to obtain a new result. But this arrangement, though convincing as an argument, in most cases affords little clue as to the way in which the construction or proof was discovered. We therefore draw the student's attention to the following hints. In attempting to solve a problem begin by assuming the required result; then by working backwards, trace the consequences of the assumption, and try to ascertain its dependence on some condition or known theorem which suggests the necessary construction. If this attempt is successful, the steps of the argument may in general be re-arranged in reverse order, and the construction and proof presented in a synthetic form. This unravelling of the conditions of a proposition in order to trace it back to some earlier principle on which it depends, is called geometrical analysis: it is the natural way of attacking the harder types of exercises, and it is especially useful in solving problems. Although the above directions do not amount to a method, they often furnish a very effective mode of searching for a suggestion. The approach by analysis will be illustrated in some of the following problems. [See Problems 23, 28, 29.] PROBLEM 20. Given a circle, or an arc of a circle, to find its centre. Let ABC be an arc of a circle whose centre is to be found. Construction. Take two chords AB, BC, and bisect them at right angles by the lines DE, FG, meeting at O. Prob. 2. Then O is the required centre. Proof. Every point in DE is equi distant from A and B. Prob. 14. And every point in FG is equidistant from B and C. .. O is the centre of the circle ABC. х Theor. 33. PROBLEM 21. To bisect a given arc. Let ADB be the given are to be bisected. Construction. Join AB, and bisect it at right angles by CD meeting the arc at D. Prob. 2. Then the arc is bisected at D. A X Then every point on CD is equidistant from A and B ; ... DA=DB; Prob. 14. :. the arcs, which subtend these angles at the O, are equal; that is, the arc DA= the arc DB. |