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5. When the circles are concentric, the centres of similitude coincide with the common centre of the circles, unless the circles are also equal, when one centre of similitude becomes any point whatever.
6. If one of the circles becomes a point, both centres of similitude coincide with the point.
288°. Def.-The circle having the centres of similitude of two given circles as end-points of a diameter is called the circle of similitude of the given circles.
The contraction ○ of s. will be used for circle of similitude. Cor. 1. Let S, S' be two circles and Z their O of s.
Since O and O' are two points from which tangents to circles S and S' are in the constant ratio of r to r', the circle o Z is co-axal with S and
S' (275°, Cor. 5).
any two circles and their of s. are co-axal.
Cor. 2. From any point P on circle Z,
Hence, at any point on the of s. of two circles, the two circles subtend equal angles.
Cor. 3. OC=CC'. ‚‚—‚, and O'C=cc'. '+r
.. I. The of s. is a line, the radical axis, when the given circles are equal (r=r').
2. The of s. becomes a point when one of the two given circles becomes a point (r or r'=0).
3. The of s. is a point when the given circles are concentric (CC'=0).
289°. Def.-With reference to the centre O (Fig. of 285°), X and Y', as also X' and Y, are called antihomologous points. Similarly with respect to the centre O', U' and Z, as also U and Y', are antihomologous points.
Let tangents at X and Y' meet at L. Then, since CX is || to C'X', ¿CXY=¿C'X'Y' =¿C'Y'X'. But ¿LXY is comp. of CXY and LLY'X' is comp. of ¿C'Y'X'.
ALXY' is isosceles, and LX-LY'.
L is on the radical axis of S and S'.
Similarly it may be proved that pairs of tangents at Y and X', at U and Y', and at U' and Z, meet on the radical axis of S and S', and the tangent at U passes through L.
... tangents at a pair of antihomologous points meet on the radical axis.
Cor. 1. The join of the points of contact of two equal tangents to two circles passes through a centre of similitude of the two circles.
Cor. 2. When a circle cuts two circles orthogonally, the joins of the points of intersection taken in pairs of one from each circle pass through the centres of similitude of the two circles.
OX: OX'=r: r',
OX. OY': OX'. OY'=r: r'.
But OX'. OY'=the square of the tangent from O to the circle S' and is therefore constant.
.. X and Y are inverse points with respect to a circle whose centre is at O and whose radius is OT'
Def. This circle is called the circle of antisimilitude, and will be contracted to O of ans.
Evidently the circles S and S′ are inverse to one another with respect to their of ans.
For the centre O' the product O'U.O'Y' is negative, and the of ans. corresponding to this centre is imaginary.
Cor. 1. Denoting the distance CC' by d, and the difference between the radii (r' − r) by d, we have
where R= the radius of the
I. When either circle becomes a point their of ans. becomes a point.
2. When the circles S and S' are equal, the of ans. becomes the radical axis of the two circles.
3. When one circle touches the other internally the of ans. becomes a point-circle. (d=d.)
4. When one circle includes the other without contact the Oof ans. is imaginary. (d<8.)
Cor. 2. Two circles and their circle of antisimilitude are co-axal. (263°)
Cor. 3. If two circles be inverted with respect to their circle of antisimilitude, they exchange places, and their radical axis being a line circle co-axal with the two circles becomes a circle through O co-axal with the two.
The only circle satisfying this condition is the circle of similitude of the two circles. Therefore the radical axis inverts into the circle of similitude, and the circle of similitude into the radical axis.
Hence every line through O cuts the radical axis and the circle of similitude of two circles at the same angle.
291°. Def.-When a circle touches two others so as to exclude both or to include both, it is said to touch them similarly, or to have contacts of like kind with the two. When it includes the one and excludes the other, it is said to touch them dissimilarly, or to have contacts of unlike kinds with the two.
292°. Theorem.--When a circle touches two other circles, its chord of contact passes through their external centre of similitude when the contacts are of like kind, and through their internal centre of similitude when the contacts are of unlike kinds.
Proof. Let circle Z touch circles S and S' at Y and X'. Then CYD and C'X'D are lines. (113°, Cor. 1)
Let XYX'Y' be the secant through Y and X'. Then
.. CX and C'X' are parallel, and X'X passes through the external centre of similitude O. (285°) Similarly, if Z' includes both S and S', it may be proved that its chord of contact passes through O.
Again, let the circle W, with centre E, touch S' at Y' and S at U so as to include S' and exclude S, and let UY' be the chord of contact. Then
..EY' and CV are parallel and VY' connects them transversely; .. VY' passes through O'.
Cor. 1. Every circle which touches S and S' similarly is cut orthogonally by the external circle of antisimilitude of S and S'.
Cor. 2. If two circles touch S and S' externally their points of contact are concyclic. (116°, Ex. 2) But the points of contact of either circle with S and S' are antihomologous points to the centre O.
.. if a circle cuts two others in a pair of antihomologous points it cuts them in a second pair of antihomologous points.
Cor. 3. If two circles touch two other circles similarly, the radical axis of either pair passes through a centre of similitude of the other pair.
For, if Z and Z' be two circles touching S and S' externally, the external circle of antisimilitude of S and S' cuts Z and Z' orthogonally (Cor. 1) and therefore has its centre on the radical axis of Z and Z'.
Cor. 4. If any number of circles touch S and S' similarly, they are all cut orthogonally by the external circle of antisimilitude of S and S', and all their chords of contact and all their chords of intersection with one another are concurrent at the external centre of antisimilitude of S and S'.
293°. Theorem.-If the circle Z touches the circles S and S', the chord of contact of Z and the radical axis of S and S' are conjugate lines with respect to the circle Z.
Proof. Let Z touch S and S' in Y and X' respectively. The tangents at Y and X' meet at a point P on the radical axis of S and S'.
But P is the pole of the chord of contact YX'.
.. the radical axis passes through the pole of the chord of contact, and reciprocally the chord of contact passes through the pole of the radical axis (267°, Def.) and the lines are conjugate.