For the equations of the three radical axes are

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

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

Ans. (-1, - 28).

*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

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, y2±2=0. These points are imaginary when we give d2 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 + d2 = 0,

therefore, since this involves the indeterminate k in the first degree, the line will always pass through the intersection of xx' +yy' + d2 = 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

The two points whose co-ordinates 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

y2 + (x − k)2 = k2 – 82,

which evidently cannot represent a real circle if k2 be less than 82; and if k2 = 8, then the equation (Art. 80) will represent a circle of infinitely small radius, the co-ordinates 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 points 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

x2+ y2 - 2kx + 8 = 0,

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

Hence, whatever be the point taken on the radical axis (¿.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.

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Ex. 3. Find the circle cutting orthogonally the three circles, Art. 108, Ex. 2. Ans. (x+1)2 + (y + 25)2 = 175748. 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' + lf" + mf''') = (k + 1 + m) c + (kc′ + lc" + 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' + IS".

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 + 2fy + c = 0,

the equation retains a single indeterminate c 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.

Ex. 7. The square of the tangent from any point of one circle to another is proportional to the perpendicular from that point upon their radical axis.

Ex. 8. To find the angle (a) at which two circles intersect.

Let the radii of the circles be R, r, and let D be the distance between their centres, then

D2 R2+2 - 2Rr cos a:

=

since the angle at which the circles intersect is equal to that between the radii to the point of intersection.

If S=0 be the equation of the circle whose radius is r, the co-ordinates of the centre of the other circle must fulfil the condition R2 - 2Rr cos a = S, as is evident from Art. 90, since D22 is the square of the tangent to S from the centre of the other circle.

Ex. 9. If we are given the angles a, ẞ at which a circle cuts two fixed circles S, S', the circle is not determined, since we have only two conditions; but we can determine the angle at which it cuts any circle of the system kS+ IS'. For we have R22Rr cos α = S, R22Rr' cos B = «S',

which is the condition that the moveable circle should cut kS+ IS' at the constant angle y: where (k + 1) r" cos y = kr cos a + lr' cos ß, r" being the radius of the circle kS+ IS'.

Ex. 10. A circle which cuts two fixed circles at constant angles will also touch two fixed circles. For we can determine the ratio k: 7 so that y shall = 0, or cos y = 1. Writing S and S' at full length and calculating the radius " of kS+IS', we easily find (k + 1) (kr2 + lr12) — klD2,

=

where D is the distance between the centres of S and S'. Substituting this value for " in the equation of the last example, we get a quadratic to determine k : 1.

113. To draw a common tangent to two circles.

We

́e saw (Art. 85) that the equation of a tangent to (S) was (x − a) (x' — a) + (y − B) (y' — B) = r2 ;

In like manner, any tangent to (S′) is

(x -- a') cos 0' + (y — B′) sin 0' = r'.

Now, if we seek the conditions necessary that these two equations should represent the same right line; first, from comparing the ratio of the coefficients of x and y, we get tan 0=tan 0',

whence ' either = 0, or = 180° +0.. If either of these conditions be fulfilled, we must equate the absolute terms, and we find, in the first case,

(a− a') cos 0+ (B — B′) sin 0 + r— r' = 0,

and in the second case,

(a−a') cose + (B − B′) sin 0 +r+r' = 0.

Either of these equations would give us a quadratic to determine . The two roots of the first equation would correspond 0.

to the direct or exterior common tangents, Aa, A'a'; the roots of the second equation would correspond to the transverse or interior tangents, Bb, B'b'.

If we wished to find the co-ordinates of the point of contact of the common tangent with the circle (S), we must substitute,

in the equation just found, for cose, its value,

x' α

and for

(a — a') (x' — a) + (B − B') (y' − B) + r (r − r') = 0 ;

or else, (a — a') (x' — a) + (B — B') (y' − B) + r (r + r') = 0.

The first of these equations, combined with the equation (S) of the circle, will give a quadratic, whose roots will be the coordinates of the points A and A', in which the direct common tangents touch the circle (S); and it will appear, as in Art. 88, that

(a' − a) (x − a) + (B' − B) (y − B) = r (r — r')

is the equation of AA', the chord of contact of direct common tangents. So, likewise,

is the equation of the chord of contact of transverse common