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duced if necessary, at D. Then the square on AC is equal to the squares on AD and CD, and the square on BĈ is equal to the squares on BD and CD (I. 47); therefore the square on AC exceeds the square on BC by as much as the square on AD exceeds the square on BD. Hence D is a fixed point either in AB or in AB produced through B (40). And the required locus is the straight line drawn through D, at right angles to AB.

50. Required the locus of a point such that the straight lines drawn from it to touch two given circles may be equal.

Let A be the centre of the greater circle, B the centre of a smaller circle; and let P denote any point on the required locus. Since the straight lines drawn from P to touch the given circles are equal, the squares on these straight lines are equal. But the squares on PA and PB exceed these equal squares by the squares on the radii of the respective circles. Hence the square on PA exceeds the square on PB, by a known square, namely a square equal to the excess of the square on the radius of the circle of which A is the centre over the square on the radius of the circle of which B is the centre. Hence, the required locus is a certain straight line which is at right angles to AB (49).

This straight line is called the radical axis of the two circles.

If the given circles intersect, it follows from III. 36, that the straight line which is the locus coincides with the produced parts of the common chord of the two circles.

51. Required the locus of the middle points of all the chords of a circle which pass through a fixed point. Let A be the centre of the given circle; B the fixed

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point; let any chord of the circle be drawn so that, produced if necessary, it may pass through B. Let P be the middle point of this chord, so that P is a point on the required focus.

The straight line AP is at right angles to the chord of which P is the middle point (III. 3); therefore P is on the circumference of a circle of which AB is a diameter. Hence if B be within the given circle the locus is the circumference of the circle described on AB as diameter; if B be without the given circle the locus is that part of the circumference of the circle described on AB as diameter, which is within the given circle.

52. O is a fixed point from which any straight line is drawn meeting a fixed straight line at P; in OP a point Q is taken such that OQ is to OP in a fixed ratio: determine the locus of Q.

We shall shew that the locus of Q is a straight line.

For draw a perpendicular from O on the fixed straight line, meeting it at C; in OC take a point D such that ŎD is to OC in the fixed ratio; draw from O any straight line OP meeting the fixed straight line at P, and in OP take a point such that OQ is to OP in the fixed ratio; join

QD. The triangles ODQ and OCP are similar (VI. 6); therefore the angle ODQ is equal to the angle OCP, and is therefore a right angle. Hence Q lies in the straight line drawn through D at right angles to OD.

53. O is a fixed point from which any straight line is drawn meeting the circumference of a fixed circle at P; in OP a point Q is taken such that CQ is to OP in a fixed. ratio: determine the locus of Q.

We shall shew that the locus is the circumference of a circle.

For let C be the centre of the fixed circle; in OC take a point D such that OD is to OC in the fixed ratio, and draw any radius CP of the fixed circle; draw DQ parallel to CP mecting OP, produced if necessary, at Q. Then the triangles OCP and ODQ are similar (VI. 4), and therefore OQ is to OP as OD is to OC, that is, in the fixed ratio. Therefore is a point on the locus. And DQ is to CP in the fixed ratio, so that DQ is of constant length. Hence the locus is the circumference of a circle of which D is the centre.

54. There are four given points A, B, C, D in a straight line; required the locus of a point at which AB and CD subtend equal angles.

Find a point in the straight line, such that the rectangle OA, OD may be equal to the rectangle OB, OC (34), and take OK such that the square on OK may be equal to either of these rectangles (II. 14): the circumference of the circle described from O as centre, with radius OK, shall be the required locus.

[We will take the case in which the points are in the following order, O, A, B, C, D.]

For let P be any point on the circumference of this

circle. Describe a circle round PAD, and also a circle

D

round PBC; then OP touches each of these circles (III. 37); therefore the angle OPA is equal to the angle PDA, and the angle OPB is equal to the angle PCB (III. 32). But the angle OPB is equal to the angles OPA and APB together, and the angle PCB is equal to the angles CPD and PDA together (I. 32). Therefore the angles OPA and APB together are equal to the angles CPD and PDA together; and the angle OPA has been shewn equal to the angle PDA; therefore the angle APB is equal to the angle CPD.

We have thus shewn that any point on the circumference of the circle satisfies the assigned conditions; we shall now shew that any point which satisfies the assigned conditions is on the circumference of the circle.

For take any point Q which satisfies the required conditions. Describe a circle round QAD, and also a circle round QBC. These circles will touch the same straight line at Q; for the angles AQB and CQD arc equal, and the converse of III. 32 is true. Let this straight line which touches both circles at Q be drawn; and let it meet the straight line containing the four given points at R. Then the rectangle RA, RD is equal to the rectangle RB, RC; for each is equal to the square on RQ (III. 36). Therefore R must coincide with O (34); and therefore RQ must be equal to OK. Thus Q must be on the circumference of the circle of which O is the centre, and OK the radius.

55. Required the locus of the vertices of all the triangles ABC which stand on a given base AB, and have the side AC to the side BC in a constant ratio.

If the sides AC and BC are to be equal, the locus is

B

the straight line which bisects AB at right angles. We will suppose that the ratio is greater than a ratio of equality; so that AC is to be the greater side.

Divide AB at D so that AD is to DB in the given ratio (VI. 10); and produce AB to E, so that AE is to EB in the given ratio. Let P be any point in the required locus; join PD and PE. Then PD bisects the angle APB, and PE bisects the angle between BP and AP produced. Therefore the angle DPE is a right angle. Therefore P is on the circumference of a circle described on DE as diameter.

We have thus shewn that any point which satisfies the assigned conditions is on the circumference of the circle described on DE as diameter; we shall now shew that any point on the circumference of this circle satisfies the assigned conditions.

Let Q be any point on the circumference of this circle, QA shall be to QB in the assigned ratio. For, take O the centre of the circle; and join QO. Then, by construction, AE is to EB as AD is to DB, and therefore, alternately, AE is to AD as EB is to DB; therefore the sum of AE and AD is to their difference as the sum of EB and DB is to their difference (23); that is, twice AO is to twice DO as twice DO is to twice BO; therefore AO is to DO as DO is

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