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It is clear that the polar of an external point must intersect the circle, and that the polar of an internal point must fall without it : also that the polar of a point on the circumference is the tangent at that point.

1. Now it has been proved (see Ex. 1, page 251] that if from an external point P two tangents PH, PK are drawn to a circle, of which O is the centre, then OP cuts the chord of contact HK at right angles at Q, so that

OP.OQ=(radius), .. HK is the polar of P with respect to the circle.

Def. (ii).

K Hence we conclude that

The Polar of an external point with reference to a circle is the chord of contact of tangents drawn from the given point to the circle.

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The following Theorem is known as the Reciprocal Property of Pole and Polar.

2. If A and P are any two points, and if the polar of A with respect to any circle passes through P, then the polar of P must pass through A.

Let BC be the polar of the point A
with respect to a circle whose centre is
O, and let BC pass through P.
Then shall the polar of p pass through A.

Join OP; and from A draw AQ perp.
to OP. We shall shew that AQ is the
polar of P.
Now since BC is the polar of A,

B :: the L ABP is a rt. angle ;

Def. (ii), p. i. and the L AQP is a rt. angle : Constr. :: the four points A, B, P, Q are concyclic; :: OQ. OP=OA. OB III. 36.

=(radius)?, for CB is the polar of A: .: P and Q are inverse points with respect to the given circle.

And since AQ is perp. to OP,

:: AQ is the polar of P.
That is, the polar of P passes through A.

Q.E.D. A similar proof applies to the case when the given point A is without the circle, and the polar BC cuts it.

3. To prove that the locus of the intersection of tangents drawn to a circle at the extremities of all chords which pass through a giver point within the circle is the polar of that point.

Let A be the given point within the circle. Let HK be any chord passing through A; and let the tangents at H and K intersect at P. It is required to prove that the locus

H of P is the polar of the point A.

I. To shew that P lies on the polar of A. Since HK is the chord of contact of

IB tangents drawn from P. :: HK is the polar of P. Ex. 1, p. ii.

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

:: the polar of A passes through P; Ex. 2, p. ii. 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. ii, 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. ii. 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.

A summary of the results obtained in this Section will be found on the following page.

SUMMARY.

(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.

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.

10. 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.

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

Fig. 2.
PI

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I. 47.

I. 47.

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 PQ-=PA2-AQ2; and PR2=PB- BRP:

· PA- AQ-= PB2 - BRP ; that is,

PS2 + AS2 - a-=PSa+SB2 – 62; or,

AS? - a2=SB2 – .
Hence AB is divided at S, so that AS2 – SB2=a? b2 :

:: 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. 1 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 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.

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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),

i.e, the radical axis of (B) and (C) passes through O. II. If the circles intersect in such a way that 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.

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