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APPENDIX.

EXAMPLES ON POLE AND POLAR.

1. Let A and B be the two given points, and let P be the intersection of their polars. Then by the Reciprocal Property of Pole and Polar, since the polar of A passes through P,

.. the polar of P passes through A.

Similarly, since the polar of B passes through P,

.. the polar of P passes through B.

Hence the polar of P passes through both A and B; that is, AB is the polar of P.

2. Let P be the intersection of the given st. lines PQ, PR, and let A and B be their poles.

Then since AB passes through A, .. its pole lies on PQ the polar of A.

Similarly since AB passes through B, .. its pole lies on PR the polar of B.

Hence the pole of AB is at P, the only point common to PQ and PR.

3. The locus must be the polar of the given point A; for by the Reciprocal Property of Pole and Polar, (i) the pole of any st. line through A must lie on the polar of A; and (ii) any point on the polar of A must be the pole of some st. line through A.

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4. Let O be the common centre, P the point of contact of any one of the tangents, and Q its pole then since the tangent is perp. to OP [III. 18], a must lie on OP (or OP produced), and OP. OQ the sq. on the radius of the given circle. But this radius is constant, and OP is constant, .. OQ is constant. Hence the locus of Q is a concentric circle.

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5. Let PQ be a diameter of one of the Os, and let o be the centre, and r the radius of the other. From O draw OT touching the first, and join OP cutting the first Join QR.

Now OR. OP=OT2 [III. 36]

= r2, since the circles are orthogonal :

and QRP is a rt. 4, being in a semicircle.

at R.

Hence QR is the polar of P: that is, the polar of P passes through Q.

6. Let P and O be the centres of the two which intersect at A, B: and let OP cut AB at Q. Join PA, PB.

Then since the Cs are orthogonal, PA and PB touch the ○ (0) at A and B hence OP. OQ = (radius)2 [Ex. 1, page 251].

And OP meets the chord of contact at rt. angles

[Ex. 2, p. 196].

.. AB is the polar of P with regard to the O (0).

O.

7. Let A and B be the given points, and O the centre of the given . Then since the polars of A and B are respectively perp. to OA, OB, .. one of the ▲ between the polars = the AQB [Ex. 3, p. 65].

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8. Let a be the point inverse to P with respect to the given Draw OY perp. to AB; and through a draw QX perp. to OP, meeting OY at X.

O.

Then since the at Q and Y are rt. angles,

.. the points Q, X, Y, P are concyclic [11. 31].

.. OX. OY = OP. OQ [111. 36]

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But OY is constant, .. OX is constant; that is, X is a fixed point.

And since the OQX is a rt. [Constr.], .. the locus of Q is a circle on Ox as diam. [111. 31].

9. Let Q be the point on OP inverse to P, and r the radius of the whose centre is O. Draw OX a diam. of the first O. Join PX, and draw QY perp. to OX.

Then OPX is a rt. ▲, being in a semicircle;

and QYX is a rt. by construction;

.. the points Q, Y, X, P are concyclic [11. 31];

.'. OX. OY = OP. OQ = r2.

But since OX is constant, .. OY is constant;

hence Y is a fixed point.

Therefore the locus of a is the st. line perp. to OX through the point inverse to X; that is, the polar of X.

10. Let C and D be the points inverse to A and B respectively, and let AX, BY be the perps. from A and B on the polars of B and A. From A and B draw AM, BN perp. respectively to OB and OA (produced if necessary).

Then

OA. OC = OB. OD = r2 [Definition].

And since the at M and N are rt. 3, the points M, B, N, A are concyclic,

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EXAMPLES ON RADICAL AXIS AND CO-AXIAL CIRCLES.

1. Let TT' be a common tangent to the two circles, and let their Radical Axis cut TT' at P. Then, by Definition, the tangential distances of the point P to the two are equal: that is, PT=PT'.

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2. Let P be any point on the Radical Axis; then the four tangents drawn from P to the two circles are equal [Def.].

Hence a described from centre P with any one of these tangents as radius will pass through all four points of contact.

And since the radii drawn from P to the points of contact are also tangents to the given circles, .. the whose centre is P cuts the given orthogonally [p. 240. Def.].

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3. As in the last example, all tangents drawn from O to the three are equal, .. a circle from centre O with radius OT will pass through all the points of contact. And since the radii of this drawn to the points of contact are also tangents to the given 3, .. the C whose centre is O cuts the given ○3 orthogonally.

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4. Let the 3 (A), (B), (C) touch one another two and two, and let OT, OT' be the common tangents of the ○3 (A), (B) and (A), (C) at their points of contact.

Then since OT and OT' are tangents to the (A),

.. OT = OT'.

That is, tangents drawn from O to the 3 (B), (C) are equal:

.. O is a point on the radical axis of the O3 (B), (C).

But the radical axis of two Os which touch one another is clearly the common tangent at their point of contact.

Hence the common tangent to the ○ (B), (C) passes also through O..

5. Take the figure of p. 243.

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Since the BEA, BEC are rt. angles, .. O described on AB, BC as diams. pass through E [III. 31];

that is, BE is the common chord of the ○ on AB and BC.

Similarly AD and CF are respectively the common chords of the s on AB, AC and on BC, CA.

Hence O, the point of intersection of the common chords, is the radical centre [p. vi.

Cor.].

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