6. See Solution of Ex. 7, p. 252. Observe that the required point B is the inverse of the given point A with regard to the given circle.

O'

S

7. Since by the last Example all which pass through the fixed point A and cut a given orthogonally, pass also through a second fixed point B (the inverse of A with regard to the given ), .. the locus of their centres is the st. line bisecting AB at rt. angles.

To find this point B, draw any radius CT to the given :

describe a to pass through A and touch CT at T [Ex. 28, p. 238]. Let this cut CA at B. Then B is the required point; for CA. CB = CT2 [111. 36].

8. Let C be the centre of the given O, and A, D the given points. Now by Ex. 6 all through A cutting the given orthogonally must pass through B the inverse point of A with respect to the given O.

Determine B as in the last Example. Then the circumscribed about the ABD is that required.

S

9. Let P be the centre of any which cuts the two given orthogonally at T and T'.

Then PT = PT', being radii.

Also PT and PT' are tangents to the given 3, since the S are cut orthogonally.

Hence the locus of P is the radical axis of the two given O

10. Let C and C' be the centres of the given C, and A the given point.

S

Then all through A cutting the (C) orthogonally pass through B the inverse of A with respect to the O (C);

and all through A cutting the (C') orthogonally pass through B' the inverse of A with respect to the (C'):

Determine the points B and B' as in the solution to Ex. 7.

Then the about the ▲ ABB' is that required.

Note that by

axis of the given

Ex. 9 the centre of this is on the radical (C) and (C').

11. Let A, B be the centres of the two given ; PQ, PR tangents to them from the given point P. Let the Radical Ax s

cut AB at S.

Draw PM, PN perp. respectively to AB and the Radical Axis ; and bisect AB at O.

12. Let A, B be the centres of two of the system, and let their Radical Axis cut AB at S. From P, any point in the Radical Axis, draw tangents PQ, PR to the two ; then PQ = PR [Hyp.]. From centre P, with radius PQ, describe a cutting AB at L, L'. Then L, L' shall be fixed points for all positions of P.

From S draw tangents ST, ST' to the two
Then

SL2 = PL2 - PS2 [1. 47]

=PQ2 - PS2

=PA2 - QA2 - PS2

But ST is independent of the position of P; .. L is a fixed

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