1.

IV. ON THE RADICAL AXIS.

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

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.

Hence AB is divided at S, so that AS - SB2= a2 - b2;

.. S is a fixed point.

I. 47.

I. 47.

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 III. 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.

B

X

Let there be three circles whose centres are A, B, C.
Let OZ be the radical axis of the * (A) and (B);

and OY the Radical Axis of the " (A) and (C), O being the point of their intersection:

then shall the radical axis of the O (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 o3 (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 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.

Let A and B be the centres of the given circles:

it is required to draw their radical axis.

If the given circles intersect, then the st. line drawn through their points of intersection will be the radical axis.

But if the given circles do not intersect,

[Ex. 1, Cor. p. 372.]

describe any circle so as to cut them in E, F and G, H:
Join EF and HG, and produce them to meet in P.

Join AB; and from P draw PS perp. to AB.
Then PS shall be the radical axis of the o3 (A), (B).

DEFINITION. If each pair of circles in a given system have the same radical axis, the circles are said to be co-axal.

EXAMPLES.

1. Shew that the radical axis of two circles bisects any one of their common tangents.

2. If tangents are drawn to two circles from any point on their radical axis; shew that a circle described with this point as centre and any one of the tangents as radius, cuts both the given circles orthogonally.

3. O is the radical centre of three circles, and from O a tangent OT is drawn to any one of them: shew that a circle whose centre is O and radius OT cuts all the given circles orthogonally.

4. If three circles touch one another, taken two and two, shew that their common tangents at the points of contact are concurrent.

5. If circles are described on the three sides of a triangle as diameter, their radical centre is the orthocentre of the triangle.

6. All circles which pass through a fixed point and cut a given circle orthogonally, pass through a second fixed point.

7. Find the locus of the centres of all circles which pass through a given point and cut a given circle orthogonally.

8. Describe a circle to pass through two given points and cut a given circle orthogonally.

9. Find the locus of the centres of all circles which cut two given circles orthogonally.

10. Describe a circle to pass through a given point and cut two given circles orthogonally.

11. The difference of the squares on the tangents drawn from any point to two circles is equal to twice the rectangle contained by the straight line joining their centres and the perpendicular from the given point on their radical axis.

12. In a system of co-axal circles which do not intersect, any point is taken on the radical axis; shew that a circle described from this point as centre with radius equal to the tangent drawn from it to any one of the circles, will meet the line of centres in two fixed points.

[These fixed points are called the Limiting Points of the system.]

13. In a system of co-axal circles the two limiting points and the points in which any one circle of the system cuts the line of centres form a harmonic range.

14. In a system of co-axal circles a limiting point has the same polar with regard to all the circles of the system.

15. If two circles are orthogonal any diameter of one is cut harmonically by the other.

Obs. In the two following theorems we are to suppose that the segments of straight lines are expressed numerically in terms of some common unit; and the ratio of one such segment to another will be denoted by the fraction of which the first is the numerator and the second the denominator.

V. ON TRANSVERSALS.

DEFINITION. A straight line drawn to cut a given system of lines is called a transversal.

1. If three concurrent straight lines are drawn from the angular points of a triangle to meet the opposite sides, then the product of three alternate segments taken in order is equal to the product of the other three segments.

Let AD, BE, CF be drawn from the vertices of the ▲ ABC to intersect at O, and cut the opposite sides at D, E, F:

then shall

BD.CE.AF=DC.EA. FB.

By similar triangles it may be shewn that

BD DC=the alt. of ▲ AOB: the alt. of ▲ AOC;

The converse of this theorem, which may be proved indirectly, is very important: it may be enunciated thus:"

If three straight lines drawn from the vertices of a triangle cut the opposite sides so that the product of three alternate segments taken in order is equal to the product of the other three, then the three straight lines are concurrent.