intersection (X) with the radical axis equal to the tangent which can be drawn from X to either circle, these points are called the limiting points of the entire system of circles which have the same (common) radical axis. They have many remarkable properties in the theory of these circles, and are such that the polar of either of them, with regard to any of the circles, is a line drawn through the other perpendicular to the line of centres. These points are real when the circles of the system have common two imaginary points, and are imaginary when they have real points common*."

When they are real it is evidently impossible for the centre of any circle of the system to lie between them, and the more nearly the centre approaches to either of them the smaller must the corresponding radius be. The limiting points themselves may therefore be considered as circles of the system of infinitely small radius.

If a system of circles have a common radical axis, and from any point on it tangents be drawn to all the circles, the locus of the points of contact must be a circle, since all these tangents are equal; and it is evident that this circle cuts any of the given system at right angles, since its radii are tangents to the given system. It is the circle passing through the limiting points of the system.

Conversely all circles which cut the given system at right angles pass through the limiting points of the system.

PROBLEM 32. (Fig. 32.) To describe a circle to touch two given circles (centres A and B, radii AD, BE respectively) and to pass through a given point C.

Take S a centre of similitude (p. 42) of A and B; draw CS and find the poles P and P, of this line with respect to each circle, (i.e. draw AP, BP, perpendicular to CS and intersecting the chords of contact of tangents from S in P and P1). Draw XR the radical axis of the given circles (p. 44): draw AC, bisect it in m and make mM on it towards C of length such that

AC AD :: AD : 2mM;

*Salmon's Conic Sections.

draw MR perpendicular to AC meeting the radical axis in R. The lines RP, RP, will cut the circles in the points of contact a, b; a, b, of the required circles and their centres can be at once found by producing Aa, Ba, &c. to meet in O and 0,.

In the figure the circles touch one of the given circles internally and one externally because S is the internal centre of similitude. If the external one be taken two more circles can be drawn, one touching both externally, the other both internally.

PROBLEM 33. To describe a circle to touch two given circles (centres A and B, radii AC, BD respectively) and a given straight line EF (fig. 33).

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Draw the radical axis of the given circles, meeting EF in R, p. 44. From A and B drop perpendiculars on the given line meeting it in E and F and the circles in C, C1, D and D, respectively. Join C,D, cutting AB in S a centre of similitude of A and B. Find P and P, the poles of this line with respect to the circles (p. 31). Draw RP, RP, cutting the circles in ab, a,b,. Then aa1, bb, are the points of contact of circles fulfilling the required conditions, and the intersections of Aa, Ba, and of Ab, Bb, give the corresponding centres. The above circles each touch both of the given circles externally or both internally since S is the external centre of similitude of A and B (p. 45). If C, be joined

to D or C to D, cutting AB in the internal centre of similitude, the poles of these lines give the points of contact of circles touching one of the given circles internally and the other externally

and if C be joined to D the poles of this line give another pair of circles touching both externally or both internally. One of these latter is shewn in the fig. There are altogether 8 solutions.

Second Solution. This problem may also be solved by dropping perpendiculars from A and B on the given line as AE, BF, bisecting the parts lying between the circles and the lines as CE, DF, in G and H and describing parabolas having A and B as foci and G and H as vertices respectively (Prob. 36). The first will necessarily be the locus of the centres of circles touching the line and the circle A externally, and the second will be the locus of the centres of circles touching the given line and the circle B externally, and hence their intersection (0) will determine the centre of a circle touching both circles externally and the given line. Similarly if C,E be bisected in G, and DF in II, and parabolas be described having A and B as foci and G1, H, as vertices respectively, each of these curves will be the

locus of centres of circles touching the line and the corresponding given circle internally. Hence the points of intersection of these four parabolas determine the centres of circles fulfilling the conditions of the problem.

The proof of the construction is obvious from the definition of a parabola subsequently given.

PROBLEM 34. To describe a circle to touch three given circles (centres ABC, radii AD, BE, CG respectively) (Fig. 34).

If the circle be required to touch the three either all externally or all internally draw the external axis of similitude SS, p. 45,

and take the poles PP,P, of this line with respect to each circle, p. 31.

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Find the radical centre R of the three circles (p. 45). Then the lines RP, RP,, RP, cut the circles in the points ab, a,b,, ab, in which the required circles must touch them: and the centre of the circle touching all three externally is given by the intersection of Aa, Ba,, Ca,, which three lines will meet in a point, and the centre of the circle touching all three internally is given by the intersection of Ab, Bb,, Cb,.

A similar construction with the remaining three axes of similitude, will determine the circles touching one internally and the remaining two externally and vice versa.

There are altogether eight solutions.

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Second solution. Join AB cutting the circles in D, D1, E and E. Bisect DE in K and DE, in K. BK will necessarily be equal to AK. With B and A as foci, and K, K, as vertices describe an hyperbola (Prob. 89), the branch of which through K will be the locus of the centres of circles touching circles A and B externally, and the branch of which through K, will be the locus of centres of circles touching these circles internally. Similarly, join BC cutting the corresponding circles in F, F, G, G. Bisect FG in L, and F,G, in L, and with C and B as foci, and L, L, as vertices, describe an hyperbola, the two branches of which will be the loci of centres of circles touching circles B and C externally and internally. The intersection of corresponding branches of the two hyperbolas will therefore determine 0,, 0,, the centres of circles touching the three given circles all externally or all internally.

Again bisecting DE, in M and DE in M, and taking B, A as foci and M, M, as vertices, an hyperbola can be described the branches of which will be the loci of centres of circles touching circles A and B, the one internally and the other externally, and the intersections of this hyperbola with that through L and L, in 0,, 0, will give centres of two more circles fulfilling the given conditions. The hyperbola through N and N1, points corresponding to

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