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4.2

SYSTEM OF TWO OR MORE CIRCLES.

spectively, which will intersect in (0) a point on AC between the given circles. The construction is obvious, since EM1 = BM = CD:

1

Common tangents to two circles may be drawn practically with all attainable accuracy by adjusting a set square to touch the circles, and drawing a line by its edge; but the points of contact should always be determined by drawing the radii perpendicular to the tangent.

Properties of a system of Two or more circles.

The points 0, 0, in which common tangents to two circles intersect are called the centres of similitude of the two circles. As is easily seen, they are the points where the line joining the centres is cut externally and internally in the ratio of the radii: and in this sense both exist when the circles cut each other, in which case of course only one pair of common tangents can be drawn, and even when one circle lies wholly inside the other, so that it is impossible to draw any common tangent.

If through a centre of similitude we draw any two lines meeting the first circle in the points R, R1, S, S1, and the second in the points p, P1, σ, σ1, then the chords RS, po will be parallel, as also the chords RS, and p11; and the chords RS and p ̧σ1, R‚§ ̧

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R1S, and po will intersect respectively in points P and Q on a line perpendicular to the line joining the centres of the circles.

This line is called the radical axis of the two circles.

The rectangle OR. OR, is constant, since it equals the square on OH the tangent from 0 (Euc. 111. 36), i. e.

and

OR. OR, OS. OS

=

Ορ. Ορ. = Οσ . Οσι.

Proof. In the triangles OAR, OCp, the angle AOR = the angle COp and OA : OC :: AR : Cp,

OR: Op :: AR: Cp (Euc. vI. 7),

.. also

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is constant and equal to the ratio of the radii

Op

of the circles wherever the line ORρ be drawn,

.. OR: OS :: Op: Oσ
Oo

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.. the triangles ROS, pOơ are similar in all respects, so that the angle ORS = the angle Opo and po is parallel to RS.

Similarly RS, is parallel to po,, which proves the first part of the proposition.

1

Again, since SRR, S, is inscribed in a circle, the angle PRO=the angle SS, R, the angle So, P. The triangles PRp, and Po,S are SS,R,

therefore similar, since the angle RPp, is common to both.

i. e.

but and

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PR. PS = square of tangent from P to circle A,
Pp1. Po1 =

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C;

.. the tangents from P to the two circles are equal, and

2

.. PA2 — AB2 – PC|2 — CD\2 ;

similarly tangents from Q to the two circles are equal.

But the locus of the intersection of equal tangents to two circles is a straight line perpendicular to the line joining their centres, and dividing the distance between them so that the difference of the squares of the parts is equal to the difference of the squares of the radii: for if X be such a point and PX perpendicular to AC, at every point on it we shall have

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2

... PA3 - PC\' = AX* — CX3 ;

and as above PA2 - PC2 = AB2 – CD2 = AX2 – C'X2.

Hence the line PQ in the figure must be such locus which proves the second part of the proposition.

DEFINITION. A line drawn perpendicular to AC, the line joining the centres of two given circles, through a point X on it, such that the difference of the squares of AX and CX is equal to the difference of the squares of the radii of the two circles is called the radical axis of the two circles.

As already shewn, it is the locus of the intersection of equal tangents to the two circles.

44

RADICAL AXIS AND AXES OF SIMILITUDE.

It may be constructed as in the last proposition or immediately from the definition by bisecting AC in a (fig. 30), and making aX towards C, the centre of the smaller circle, a fourth proportional 2AC, AB+CD, and AB – CD,

to

i.e. by making

aX : R-r:
−r :: R+r : 2AC,

where R and r are the radii of the circle, and drawing a line through X perpendicular to AC; for in this case

AC × 2αX = R2 – r2 but AC = AX + XC and 2aX = AX − CX

··. (AX+CX) (AX − CX) = R2 — r2 = AX2 – CX2.

The radical axis bisects the distance between the polars with respect to the two circles, of either centre of similitude, which furnishes another method of constructing it.

Given three circles (centres C, C1, C,, radii r, r1, "1⁄2); the line joining a centre of similitude of C and C, to a centre of similitude of C and C, will pass through a centre of similitude of C, Let S and S, (fig. 31) be the centres of similitude of

and C'2.

2

Fig.31.

S

C and C1, and S1 a centre of similitude of C and C2, and let SS1, C1C, meet in S, S will be a centre of similitude of C, and C.

2

For since

1

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or CSCS, is a harmonic range; therefore S (CS'C1S) is a harmonic pencil, and therefore if CC, cuts SS,' in S',

2

CSCS, is a harmonic range, and since S is a centre of

1

similitude of C and C2, S must be the other.

Through C, draw CL parallel to CS, and meeting SS in L.

2

2

Then by similar triangles

2

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or S is a centre of similitude of C, and C2.

Since for each pair of circles there are two centres of similitude, there will be in all six for the three circles, and these will be distributed along four axes of similitude, as represented in the figure.

Corollary. If a circle (centre A) touch two others (centres C and C1) the line joining the points of contact will pass through a centre of similitude of C and C1. For when two circles touch, one of their centres of similitude will coincide with the point of contact. If A touch C and C1 either both externally or both internally, the line joining the points of contact will pass through the external centre of similitude of C and C1. If A touch one externally and the other internally, the line joining the points of contact will pass through the internal centre of similitude*.

1

1

Given any three circles, if we take the radical axis of each pair of circles, these three lines will meet in a point, which is called the radical centre of the three circles.

For let the radical axes of A and C and of B and C intersect in R (fig. 34), then the tangents from R to A and C are equal, as also the tangents from R to B and C; therefore the tangent from R to A must be equal to the tangent from R to B, i. e. R must be a point on the radical of A and B, which proves the proposition.

If two circles have a common radical axis, and points L and L ̧ be taken on the line joining their centres at a distance from its

* Salmon's Conic Sections.

46

LIMITING POINTS OF A SYSTEM OF CIRCLES.

intersection (I) 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.

1

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.

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