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CHAPTER VIII.

PROPERTIES OF A SYSTEM OF TWO OR MORE CIRCLES.

105. To find the equation of the chord of intersection of two circles.

If S=0, S'=0 be the equations of two circles, then any equation of the form S+kS'0 will be the equation of a figure passing through their points of intersection (Art. 40).

Let us write down the equations

8 = (x − a )* + (y − B )3 — p3 = 0,

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and it is evident that the equation S+kS'=0 will in general represent a circle, since the coefficient of xy=0, and that of x2= that of y3. There is one case, however, where it will represent a right line, namely, when k-1. The terms of the second degree then vanish, and the equation becomes

S- S'= 2 (α' – a) x + 2 (B′ − B) y + r22 — r2 + a2 — a” + ß3 − ß”= 0.

This is, therefore, the equation of the right line passing through the points of intersection of the two circles.

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What has been proved in this article may be stated as in Art. 50. If the equation of a circle be of the form S+kS' 0 involving an indeterminate k in the first degree, the circle passes through two fixed points, namely, the two points common to the circles S and S'.

106. The points common to the circles S and S' are found by seeking, as in Art. 82, the points in which the line S-S' meets either of the given circles. These points will be real, coincident, or imaginary, according to the nature of the roots of the resulting equation; but it is remarkable that, whether the circles meet in real or imaginary points, the equation of the chord of intersection, S-S=0, always represents a real line, having important geometrical properties in relation to the two circles. This is in conformity with our assertion (Art. 82), that

the line joining two points may preserve its existence and its properties when these points have become imaginary.

In order to avoid the harshness of calling the line S-S', the chord of intersection in the case where the circles do not geometrically appear to intersect, it has been called* the radical axis of the two circles.

107. We saw (Art. 90) that if the coordinates of any point ay be substituted in S, it represents the square of the tangent drawn to the circle S from the point xy. So also S' is the square of the tangent drawn to the circle S'; hence the equation S-S'0 asserts, that if from any point on the radical axis tangents be drawn to the two circles, these tangents will be equal.

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The line (S-S') possesses this property whether the circles meet in real points or not. When the circles do not meet in real points, the position of the radical axis is determined geometrically by cutting the line joining their centres, so that the difference of the squares of the parts may the difference of the squares of the radii, and erecting a perpendicular at this point; as is evident, since the tangents from this point must be equal to each other.

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If it were required to find the locus of a point whence tangents to two circles have a given ratio, it appears, from Art. 90, that the equation of the locus will be S-k'S'=0, which (Art. 105) represents a circle passing through the real or imaginary points of intersection of S and S'. When the circles S and S' do not intersect in real points, we may express the relation which they bear to the circle S-k'S', by saying that the three circles have a common radical axis.

Ex. Find the coordinates of the centre, and the radius of kS + IS'.

kala' kẞ+ IB Ans. Coordinates are ; that is to say, the line joining the centres k+ l k + l of S, S' is divided in the ratio : 1. Radius is given by the equation (k + 1)2 p''2 = (k + 1) (kr2 + lr'2)

klD2,

where D is the distance between the centres of S and S'.

108. 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.

* By M. Gaultier, of Tours (Journal de l'École Polytechnique, Cahier XVI. 1813).

For the equations of the three radical axes are

S-S' 0, S'-S" =0, S" - S=0,

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which, by Art. 41, meet in a point.

From this theorem we immediately derive the following: If several circles pass through two fixed points, their chords of intersection with a fixed circle will pass through a fixed point.

For, imagine one circle through the two given points to be fixed, then its chord of intersection with the given circle will be fixed; and its chord of intersection with any variable circle drawn through the given points will plainly be the fixed line joining the two given points. These two lines determine by their intersection a fixed point through which the chord of intersection of the variable circle with the first given circle must pass.

Ex. 1. Find the radical axis of

x2 + y2 — 4x − 5y + 7 = 0; x2 + y2 + 6x + 8y − 9 = 0.

Ex. 2. Find the radical centre of

Ans. 10x+13y = 16.

(x − 1)2 + (y − 2)2 = 7; (x − 3)3 + y2 = 5; (x+4)2 + (y + 1)2 = 9.

Ans. (--18).

*109. A system of circles having a common radical axis possesses many remarkable properties, which are more easily investigated by taking the radical axis for the axis of y, and the line joining the centres for the axis of x. Then the equation of any circle will be

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where is the same for all the circles of the system, and the equations of the different circles are obtained by giving different values to k. For it is evident (Art. 80) that the centre is on the axis of x, at the variable distance k; and if we make x = 0 in the equation, we see that no matter what the value of k may be, the circle passes through the fixed points on the axis of y, y*+8=0. These points are imaginary when we give & the sign+, and real when we give it the sign

*110. The polars of a given point, with regard to a system of circles having a common radical axis, always pass through a fixed point.

The equation of the polar of x'y' with regard to

x2 + y2 − 2kx + 8 = 0,

is (Art. 89)

xx′+yy − k (x + x′) + 82 = 0 ;

therefore, since this involves the indeterminate k in the first degree, the line will always pass through the intersection of xx2+yy' +8=0, and x+x=0.

*111. There can always be found two points, however, such that their polars, with regard to any of the circles, will not only pass through a fixed point, but will be altogether fixed.

This will happen when xx+yy' + 8 = 0 and x+x′ = 0 represent the same right line, for this right line will then be the polar whatever the value of k. But that this should be the case. we must have

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The two points whose coordinates have been just found 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 imaginary when they have real points common.

The equation of the circle may be written in the form

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which evidently cannot represent a real circle if k* be less than 82; and if k2=82, then the equation (Art. 80) will represent a circle of infinitely small radius, the coordinates of whose centre are y=0, x=8. Hence the points just found may themselves be considered as circles of the system, and have, accordingly, been termed by Poncelet* the limiting points of the system of circles.

*112. If from any point on the radical axis we draw tangents to all these circles, the locus of the point of contact must be a circle, since we proved (Art. 107) that all these tangents were equal. It is evident, also, that this circle cuts any of the given system at right angles, since its radii are tangents to the given system. The equation of this circle can be readily found.

Traité des Propriétés Projectives, p. 41.

The square of the tangent from any point (x=0, y=h) to the circle

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being found by substituting these coordinates in this equation is h+8; and the circle whose centre is the point (x = 0, y = h), and whose radius squared = h + 8, must have for its equation x* + (y − h)* = h2 + 82,

or

x2 + y2 - 2hy = 82.

Hence, whatever be the point taken on the radical axis (i.e. whatever the value of h may be), still this circle will always pass through the fixed points (y=0, x=±8) found in the last Article. And we infer that all circles which cut the given system at right angles pass through the limiting points of the system.

Ex. 1. Find the condition that two circles

x2 + y2 + 2gx + 2ƒy + c = 0, x2 + y2 + 2g'x + 2ƒ'y + c′ = 0 should cut at right angles. Expressing that the square of the distance between the centres is equal to the sum of the squares of the radii, we have

or, reducing,

(g−g')2 + (ƒ − ƒ ')2 = g2 + ƒ2 − c + g'2 + ƒ12 − c',
2gg' +2ff' = c + c'.

Ex. 2. Find the circle cutting three circles orthogonally. We have three equations of the first degree to determine the three unknown quantities g, f, c; and the problem is solved as in Art. 94. Or the problem may be solved otherwise, since it is evident from this article that the centre of the required circle is the radical centre of the three circles, and the length of its radius equal to that of the tangent from the radical centre to any of the circles.

Ex. 3. Find the circle cutting orthogonally the three circles, Art. 108, Ex. 2.
Ans. (x+)2 + (y + 78)2 = 178.

Ex. 4. If a circle cut orthogonally three circles S', S", S"", it cuts orthogonally any circle kS' + IS" + mS"" = 0. Writing down the condition

2g (kg′ + lg" + mg") + 2ƒ ( kf' + lƒ" + mƒ"'') = (k + 1 + m) c + (kc′ + le" + mc'''), we see that the coefficients of k, l, m vanish separately by hypothesis.

Similarly, a circle cutting S, S" orthogonally, also cuts orthogonally kS' + 18′′.

Ex. 5. A system of circles which cuts orthogonally two given circles S', S" has a common radical axis. This, which has been proved in Art. 112, may be proved otherwise as follows: The two conditions

2gg' + 2ff' = c + c', 2gg" + 2ff" =c+c",

enable us to determine g and ƒ linearly in terms of c. Substituting the values so found in

x2 + y2+2gx + 2ƒy + c = 0,

the equation retains a single indeterminate e in the first degree, and therefore (Art. 105) denotes a system having a common radical axis.

Ex. 6. If AB be a diameter of a circle, the polar of A with respect to any circle which cuts the first orthogonally will pass through B.

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