Conversely, to determine Y, describe any circle through P and P1, let it cut K in A and B, produce AB to meet PP1 in M, and through M draw a tangent to K. When Y is known, the problem is reduced to Ex. 1. Fig. 83 shows the construction.

Discussion. The two tangents through M to K furnish, in general, two solutions. Special cases: (i.) P without K, and P1 in the tangent from P to K; (ii.) P in the circumference of K and P1, (a) in, (b) within, (c) without, the circumference of K; (iii.) P and P1 within K; (iv.) P, P1, and O in a straight line; (v.) P and P1 equidistant from K.

M

B

Fig. 83.

4. To construct a circle through P, and touching L, L1.

Analysis. Bisect that angle formed by L and L1 (Fig. 84), in which Plies; draw a perpendicular PB from P to the bisector; produce PB to Q, making BQ = BP; then Q is a second point in the required circumference, and the problem is reduced to Ex. 2.

Second Analysis. About any point O (Fig. 84), in the bisector as centre, describe a circle touching L and L1; the point A is a centre of similitude of this circle and the required circle. (Ex. 36, % 13.) Let AP cut this auxiliary circle in D and E; then OD or OE is parallel to the radius of the required circle drawn to P. (Ex. 33, 13.)

Discussion. In general, there are two solutions. Special cases: (i.) L || to L1 ; (ii.) Plies in L; (iii.) Plies in the bisector AQ.

5. To construct a circle through P, and touching L, K.

Analysis. Let the required circle touch Lin Y (Fig. 85), and K in Z. Produce YZ to meet K in A, join AO, and produce it to meet K a second time in B and L in C. Join BZ and AP, and let AP meet the required circle a second time in Q.

LXYZ=2XZY= LAZO = LOAZ. .. AC is to XY. But XY is 1 to L; .. AC is 1 to L. Also, BZ is to AY; .. the AACY, ABZ are similar; whence, AB × AC = AZ× AY. Also, AQ× AP - AZ× AY (No. 165). Hence, ABX AC= AQ× AP. .. the circumference which passes through B, C, and P must also pass through Q. (Ex. 10, § 13.) By describing this circumference, Q is found, and the problem is then reduced to Ex. 2.

=

Discussion. If L does not cut K, and Plies without K, as shown in Fig. 85, there are four solutions, two of which are obtained by joining P to A, and the other two by joining P to B. There are numerous special cases for different relative positions of P, L, and K.

=

Fig. 86.

=

6. To construct a circle through P, and touching K, K1. Analysis. Let the required circle touch K, K, in Y, Z (Fig. 86). Produce YZ to meet 00, in M. LOYB = L XYZ = 2 XZY: Z AZO1 201AZ, .. OY is | to O1A, .. M is the direct centre of similitude of K and K1 (Ex. 32, § 13), .. MD × MC = MY × MZ (Ex. 35, 13). Join PM, cutting the required circle a second time in Q; then MPX MQ=MY× MZ(No. 165), .. MP× MQ=MD× MC, whence it follows that Qlies in the circumference passing through the three known points P, D, C. By describing this circumference and joining P to M, the direct centre of similitude of K and K1, the point Q is found, and the problem will then be reduced to Ex. 3.

Discussion. The maximum number of solutions is four. In discussing the numerous special cases, the position of P relatively to the common tangents of K and K1 must be considered.

7. To construct a circle touching L, L1, L.

See Ex. 66, 7. According to the relative positions of the lines the number of solutions is four, two, or none.

8. To construct a circle touching L, L1, K.

Analysis. Suppose the problem solved, and a concentric circle described with X as centre and XO as radius (Fig. 87); this circle must touch parallels to L and L1, respectively, drawn at the distance r from L and L1. Therefore X is found by constructing this concentric circle, as explained in Ex. 4.

Discussion. The maximum number of solutions is 8. In discussing special cases, the positions of L and L1 relatively to K must be considered.

S. To construct a circle touching L, K, K1.

Analysis. Suppose the required circle constructed, and also a concentric circle with X as centre and XO as radius; this circle must touch : (i.) a parallel to L, drawn at the distance r from L; (ii.) a circle with O, as centre and r1-r as radius. Hence, X is found by constructing this concentric circle, as explained in Ex. 5.

Discussion. The use of r1 r as radius of the auxiliary circle may give four solutions, and the use of r1+r four more solutions.

10. To construct a circle touching K, K1, K.

Analysis. A concentric circle, with XO as radius, must touch the circle with O as centre and r1 -r as radius, and also touch the circle with O, as centre and r2 r as radius. By constructing this concentric circle, therefore, as explained in Ex. 6, X will be determined.

2

Discussion. By using for the radii of the two auxiliary circles about O1 and O2 as centres different combinations of the four values, r1 −r, r1 +r, r2 −r, r2 +r, the maximum number of solutions is eight.

CHAPTER IV.

EQUIVALENT FIGURES.

§ 19. THEOREMS.

1. All parallelograms having equal bases, and contained between two parallel lines, are equivalent.

2. Two triangles having two sides equal, each to each, and the included angles supplementary, are equivalent.

3. Two parallelograms having two adjacent sides equal, each to each, and the included angles supplementary, are equivalent.

4. Every straight line drawn through the centre of a parallelogram divides it into two equivalent parts. What kind of figures are the two parts?

5. Every straight line drawn through the middle point of the median of a trapezoid, and cutting the two bases, divides the trapezoid into two equivalent parts. Is the theorem also true if the line cuts the legs instead of the bases?

6. In every trapezoid the triangle which has for base one of the legs of the trapezoid, and for vertex the middle point of the opposite side, is equivalent to one-half of the trapezoid.

7. The sum of the areas of two opposite triangles, formed by joining a point within a parallelogram to the four vertices, is equal to one-half of the area of the parallelogram.