2. If the circles be unequal, and the line be required to touch them on the same side of the line joining the centres. Let A and B be the centres; join AB; and with the D E centre B and distance equal to the difference of the given radii, describe a circle, and from A draw AE touching it. Join BE and produce it to D, draw AC parallel to BD, and join CD. Then AC being parallel and equal to DE, CD is equal and parallel to AE, .. ACDE is a parallelogram ; and the angle AEB being a right angle, AED is also a right angle; hence the angles at C and D are right angles, and therefore CD touches both circles. 3. If the line be required to touch them on opposite sides of the line joining the centres. With the centre B and radius equal to the sum of B the given radii describe a circle, to which from A draw a tangent AE. Join BE, and let it cut the given circle in D. Draw AC parallel to BE; join CD. Then AC being equal and parallel to ED, ACDE is a parallelogram, and the angle AED being a right angle, the angles at C and D are right angles, and therefore CD touches both circles. (8.) If a line touching two circles cut another line joining their centres, the segments of the latter will be to each other as the diameters of the circles. Let the line AB touch the circles, whose centres are C and D, in A and B, and cut CD in the point E; CE will be to ED in the ratio of the diameters of the circles. E B Join CA, BD. Then the angles at A and B are right angles and the angles at E are vertically opposite, therefore the triangles AEC, BED are equiangular, and consequently CE: ED :: CA : BD :: 2 CA: 2BD. (9.) If a straight line touch the interior of two concentric circles, and be placed in the outer; it will be bisected at the point of contact. Let AB touch the interior of two circles, whose common centre is O, in the point C; AB is bisected in C. Join OC; then (Eucl. iii. 18.) the angles at Care right angles; and OC' drawn A D B C from the centre of the circle ADB at right angles to AB, bise cts it (Eucl. iii. 3.). GEOMETRICAL PROBLEMS. [Sect. 2. (10.) If any number of equal straight lines be placed in a circle; to determine the locus of their points of bisection. E D B Let there be any number of lines AB, CD, placed in the circle whose centre is O, and let them be bisected in E, F; join OE, OF; then (Eucl. iii. 14.) these lines are equal, and therefore the locus will be a circle whose centre is O, and radius equal to the distance of the points of bisection from 0. (11.) If from a point in the circumference of a circle, any number of chords be drawn; the locus of their points of bisection will be a circle. From the given point A let any chord AB be drawn in the circle, whose centre is 0; bisect it in D. Join AO, BO and draw DE parallel to BO. A E B Then DE being parallel to BO, the triangles ADE, ABO are similar, and BO is equal AO, .. DE=EA; but AE: A0 :: AD: AB (Eucl. vi. 2.), whence AE = AO, .. ED= EA = AO, and the locus will be a circle described on ДO as a diameter. (12.) If on the radius of a given semicircle, another semicircle be described, and from the extremity of the diameters any lines be drawn cutting the circumferences, and produced so that the part produced may always have a given ratio to the part intercepted between the two circumferences; to determine the locus of the extremities of these lines. On AB the radius of the semicircle AEC let a semicircle ADB be described; and from A draw any line ADE, which produce till EF: ED in the given ratio. B F E Produce AC to G, making CG: CB, in the given ratio, and join DB, EC, FG; then since FE: ED :: : GC : CB, .. FE GC :: ED: CB :: DA : AB :: EA: CA, whence (Eucl. vi. 2.) FG is parallel to CE and DB, and the angle AFG is a right angle, and is in a semicircle whose diameter is AG; hence the locus required is a semicircle. (13.) If from a given point without a given circle, straight lines be drawn, and terminated by the circumference; to determine the locus of the points which divide them in a given ratio. that AO: AE in the given ratio, and find a point F, so that EF may be to OD in the given ratio, and with the centre E and radius EF describe a circle; it will be the locus required. Draw any line AGC; join OC, EG. Since 40: AE in a given ratio, as also OD: EF; .. OC: EG :: AO : AE, X hence OC is parallel to EG, and AC: AG: OC: EG, i. e. in the given ratio. In the same manner it may be shewn that every line drawn from A to BCD will be divided by the circumference of the circle GFH in the same ratio, i. e. GFH will be the locus required. (14.) Having given the radius of a circle; to determine its centre, when the circle touches two given lines, which are not parallel. Let BA, AC be the two lines which touch the circle, whose radius is given. Bisect the angle BAC by the line AE, the centre of the circle will be in this line (Eucl. iv. 4.) D A B E From A draw AD at right angles to AB and make it equal to the given radius; through D draw DO parallel to AB meeting AE in O; then the centre of the circle being in this line also, must be at the point of intersection O. (15.) Through three given points which are not in the same straight line, a circle may be described; but no other circle can pass through the same points. Let A, B, C be the three given points. Join AB, BC, and bisect them in D and E; from which points draw DO, EO at right angles to them; these lines will meet in some point 0; for if not, they are F H O E مم D B |