Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

But HK, the polar of P, passes through A;

:: the polar of A passes through P: Ex. 2, p. 392. that is, the point P lies on the polar of A.

II. To shew that any point on the polar of A satisfies the given conditions.

Let BC be the polar of A, and let P be any point on it.
Draw tangents PH, PK, and let HK be the chord of contact.

Now from Ex. 1, p. 391, we know that the chord of contact HK is the polar of P, and we also know that the polar of P must pass through A; for P is on BC, the polar of A:

Ex. 2, p. 392. that is, HK passes through A. :: P is the point of intersection of tangents drawn at the extremities of a chord passing through A.

From I. and II. we conclude that the required locus is the polar of A.

NOTE. If A is without the circle, the theorem demonstrated in Part I. of the above proof still holds good; but the converse theorem in Part II. is not true for all points in BC. For if A is without the circle, the polar BC will intersect it; and no point on that part of the polar which is within the circle can be the point of intersection of tangents.

We now see that

(i) The Polar of an external point with respect to a circle is the chord of contact of tangents drawn from it.

(ii) The Polar of an internal point is the locus of the intersections of tangents drawn at the extremities of all chords which pass through it.

(iii) The Polar of a point on the circumference is the tangent at that point.

The following theorem is known as the Harmonic Property of Pole and Polar.

4. Any straight line drawn through a point is cut harmonically by the point, its polar, and the circumference of the circle.

Let AHB be a circle, p the given point and HK its polar; let Paqb be any

aight line drawn through P meeting the polar at q and the Oce of the circle at a and b. Then shall P, a, q, b be a harmonic

A

B range.

In the case here considered, P is an external point.

K K
Join P to the centre O, and let PO
cut the Oce at A and B: let the polar of
P cut the Oce at H and K, and PO at Q.

Join Qa, Qb, Oa, OH, Ob, PH.

Then PH is a tangent to the O AHB. Ex. 1, p. 391. From the similar triangles OPH, HPQ,

OP : PH=PH : PQ. .:. PQ . PO=PHP

=Pa. Pb. :: the points O, Q, a, b are concyclic: .: the LaQA=the L abo Ex. 5, p. 241.

=the L Oab

=the LOQb, in the same segment. And since QH is perp. to AB,

:. the L aQH=the I QH. :: Qq and QP are the internal and external bisectors of the L aQb:

:: P, a, q, b is a harmonic range. Ex. 1, p. 385. The student should investigate for himself the case when P is an internal point.

Conversely, it may be shewn that if through a fixed point P any secant is drawn cutting the circumference of a given circle at a and b, and if q is the harmonic conjugate of P with respect to a, b; then the locus of q is the polar of P with respect to the given circle.

I. 5.

DEFINITION.

A triangle so

ela to a that each side is the polar of the opposite vertex is said to be self-conjugate with respect to the circle.

EXAMPLES ON POLE AND POLAR.

1. The straight line which joins any two points is the polar with respect to a given circle of the point of intersection of their polars.

2. The point of intersection of any two straight lines is the pole of the straight line which joins their poles.

3. Find the locus of the poles of all straight lines which pass through a given point.

4. Find the locus of the poles, with respect to a given circle, of tangents drawn to a concentric circle.

5. If two circles cut one another orthogonally and PQ be any diameter of one of them ; shew that the polar of P with regard to the other circle passes through Q.

6. If two circles cut one another orthogonally, the centre of each circle is the pole of their common chord with respect to the other circle.

7. Any two points subtend at the centre of a circle an angle equal to one of the angles formed by the polars of the given points.

8. O is the centre of a given circle, and AB a fixed straight line.

P is any point in AB ; find the locus of the point inverse to P with respect to the circle.

9. Given a circle, and a fixed point O on its circumference : P is any point on the circle. find the locus of the point inverse to p with respect to any circle whose centre is O.

Given two points A and B, and a circle whose centre is O; shew that the rectangle contained by OA and the perpendicular from B on the polar of A is equal to the rectangle contained by OB and the perpendicular from A on the polar of B.

11. Four points A, B, C, D are taken in order on the circumference of a circle ; DA, CB intersect at P, AC, BD at Q, and BA, CD in R: shew that the triangle PQR is self-conjugate with respect to the circle.

12. Give a linear construction for finding the polar of a given point with respect to a given circle. Hence find a linear construction for drawing a tangent to a circle from an external point.

13. If a triangle is self-conjugate with respect to a circle, the centre of the circle is at the orthocentre of the triangle.

14. The polars, with respect to a given circle, of the four points of a harmonic range form a harmonic pencil : and conversely.

10.

[blocks in formation]

1. To find the locus of points from which the tangents drawn to two given circles are equal.

[blocks in formation]
[ocr errors]

Let A and B be the centres of the given circles, whose radii are a and b; and let P be any point such that the tangent PQ drawn to the circle (A) is equal to the tangent PR drawn to the circle (B).

It is required to find the locus of P.
Join PA, PB, AQ, BR, AB; and from P draw PS perp. to AB.

Then because PQ=PR, PQ2 = PR?.
But PQ2=PA2 – AQ2; and PR2=PBP - BR2 :

1. 47. :: PAP - AQ= PB2 – BR2; that is, PS2+AS- a2= PS2 + SB2 – 12;

I. 47. AS2 – a2=SB2 - 62. Hence AB is divided at S, so that ASP – SB=a? 62:

:. S is a fixed point. Hence all points from which equal tangents can be drawn to the two circles lie on the straight line which cuts AB at rt. angles, so that the difference of the squares on the segments of AB is equal to the difference of the squares on the radii.

Again, by simply retracing these steps, it may be shewn that in Fig. I every point in SP, and in Fig. 2 every point in SP exterior to the circles, is such that tangents drawn from it to the two circles are

or,

equal.

Hence we conclude that in Fig. 1 the whole line SP is the required locus, and in Fig. 2 that part of SP which is without the circles.

In either case SP is said to be the Radical Axis of the two circles.

COROLLARY. If the circles cut one another as in Fig. 2, it is clear that the Radical Axis is identical with the straight line which passes through the points of intersection of the circles ; for it follows readily from 111. 36 that tangents drawn to two intersecting circles from any point in the common chord produced are equal.

2. The Radical Axes of three circles taken in pairs are concurrent.

[merged small][ocr errors][merged small][merged small][ocr errors]

Let there be three circles whose centres are A, B, C.

Let OZ be the radical axis of the OS (A) and (B); and OY the Radical Axis of the OS (A) and (C), O being the point of their intersection.

Then shall the radical axis of the Os (B) and (C) pass through O.

It will be found that the point O is either without or within all the circles.

I. When O is without the circles.

From O draw OP, OQ, OR tangents to the OS (A), (B), (C). Then because O is a point on the radical axis of (A) and (B); Hyp.

.. OP=OQ. And because O is a point on the radical axis of (A) and (C), Hyp.

.. OP=OR;

:: OQ=OR; O is a point on the radical axis of (B) and (C); that is, the radical axis of (B) and (C) passes through O. II. If the circles intersect in such a way that O is within them all; the radical axes are then the common chords of the three circles taken two and two; and it is required to prove that these common chords are concurrent. This may be shewn indirectly by III. 35.

DEFINITION. The point of intersection of the radical axes of three circles taken in pairs is called the radical centre.

« ΠροηγούμενηΣυνέχεια »