and for the inverse pair; and, therefore, so is also the ratio to rr' of D2 - (r±r').

Consider now four circles touching the same right line in four points. Now the mutual distances of four points on a right line are connected by the relation 12.34+14.3213.21; as may easily be proved by the identical equation

(b − a) (d − c) + (d − a) (c — b) = (c − a) (d — b),

where a, b, c, d denote the distances of the points from any origin on the line. Thus then the common tangents of four circles which touch the same right line are connected by the relation which is to be proved. But if we take the inverse of the system with regard to any point, we get four circles touched by the same circle, and the relation subsists still; for if the equation be divided by the square root of the products of all the radii, it consists of members

12

34

√ (2'x')' √(x'j''') which are unchanged by the process of inversion.

&c.,

The relation between the common tangents being proved in this way, we have only to suppose the four circles to become four points, when we deduce as a particular case the relation connecting the sides and diagonals of an inscribed quadrilateral. This method also shews that, in the case of two circles which touch the same side of the enveloping circle, we are to use the direct common tangent; but the transverse common tangent when one touches the concavity, and the other the convexity of that circle. Thus then we get the equation of the four pairs of circles which touch three given circles,

23 √(S) ± 31 √(S′) ± 12 √/ (S”) = 0.

When 12, 23, 31 denote the lengths of the direct common tangents, this equation represents the pair of circles having the given circles either all inside or all outside. If 23 denotes a direct common tangent, and 31, 12 transverse, we get a pair of circles each having the first circle on one side, and the other two on the other. And, similarly, we get the other pairs of circles by taking in turn 31, 12 as direct common tangents, and the other common tangents transverse.

* Another proof will be given in the appendix to the next chapter.

*CHAPTER IX.

APPLICATION OF ABRIDGED NOTATION TO THE EQUATION
OF THE CIRCLE.

122. IF we have an equation of the second degree expressed in the abridged notation explained in Chap. IV., and if we desire to know whether it represents a circle, we have only to transform to x and y coordinates, by substituting for each abbreviation (a) its equivalent (x cos a+y sina-p); and then to examine whether the coefficient of xy in the transformed equation vanishes, and whether the coefficients of x2 and of y' are equal. This is sufficiently illustrated in the examples which follow.

When will the locus of a point be a circle if the product of perpendiculars from it on two opposite sides of a quadrilateral be in a given ratio to the product of perpendiculars from it on the other two sides?

Let α, B, Y, ♪ be the four sides of the quadrilateral, then the equation of the locus is at once written down ay=kßd, which represents a curve of the second degree passing through the angles of the quadrilateral, since it is satisfied by any of the four suppositions,

a=0, B=0; a=0, 8=0; B=0, y = 0; y = 0, d=0. Now, in order to ascertain whether this equation represents a circle, write it at full length

=

(x cosa + y sina − p) (x cosy + y sin y—p")

k (x cosß + y sin ẞ - p') (x cos♪ + y sin d − p"). Multiplying out, equating the coefficient of x to that of y*, and putting that of xy = 0, we obtain the conditions

cos (a+y)= cos(B+8); sin (a + y) = k sin (B+8). Squaring these equations, and adding them, we find k=±1; and if this condition be fulfilled, we must have

whence

a+y=ẞ+8, or else = 180°+B+d;

a-ß=d-y, or 180+8-y.

Recollecting (Art. 61) that a-ẞ is the supplement of that angle between a and B, in which the origin lies, we see that this condition will be fulfilled if the quadrilateral formed by aßyd be inscribable in a circle (Euc. III. 22). And it will be seen on examination that when the origin is within the quadrilateral we are to take k=-1, and that the angle (in which the origin lies) between a and B is supplemental to that between y and 8; but that we are to take k=+1, when the origin is without the quadrilateral, and that the opposite angles are equal.

123. When will the locus of a point be a circle, if the square of its distance from the base of a triangle be in a constant ratio to the product of its distances from the sides?

Let the sides of the triangle be a, B, y, and the equation of the locus is aẞky. If now we look for the points where the line a meets this locus, by making in it a=0, we obtain the perfect square y2=0. Hence a meets the locus in two coincident points, that is to say (Art. 83), it touches the locus at the point ay. Similarly, B touches the locus at the point By. Hence a and ẞ are both tangents, and y their chord of contact. Now, to ascertain whether the locus is a circle, writing at full length as in the last article, and applying the tests of Art. 80, we obtain the conditions

cos (a + B) = k cos 2y; sin (a+B) = k sin 2y; whence (as in the last article) we get k=1, a-y=y-ẞ, or the triangle is isosceles. Hence we may infer that if from any point of a circle perpendiculars be let fall on any two tangents and on their chord of contact, the square of the last will be equal to the rectangle under the other two.

Ex. When will the locus of a point be a circle if the sum of the squares of the perpendiculars from it on the sides of any triangle be constant ?

The locus is a2 + ß2 + y2 = c2; and the conditions that this should represent a circle are

cos 2a + cos 28+ cos 2y = 0;

cos 2α = - 2 cos (B+ y) cos (3-y);

Squaring and adding.

sin 2a + sin 28 + sin 2y = 0.

sin 2a = - 2 sin (6+ y) cos (3 - y).

1 = 4 cos2 (6-y); ẞ — y = 60°.

And so, in like manner, each of the other two angles of the triangle is proved to be 60°, or the triangle must be equilateral.

124. To obtain the equation of the circle circumscribing the triangle formed by the lines a=0, B = 0, y = 0.

Any equation of the form

Iẞy + mya+naß = 0

denotes a curve of the second degree circumscribing the given triangle, since it is satisfied by any of the suppositions

a=0, B=0; B=0, y=0; y = 0, a=0.

The conditions that it should represent a circle are found, by the same process as in Art. 122, to be

l cos (B+ y) +m cos (y + a) +n cos (a + B) = 0,

7 sin (B+ y) +m sin (y+a) + n sin (a + B) = 0.

Now we have seen (Art. 65) that when we are given a pair of equations of the form

la' + mß' + ny' = 0, la" + mß" + ny′′ = 0,

1, m, n must be respectively proportional to B'y"-B'y', y'a"—y'a', a'ß" - a"ß'. In the present case then l, m, n must be proportional to sin (ß − y), sin (y — a), sin (a− ß), or (Art. 61) to sin A, sin B, sin C.

scribing a triangle is

Hence the equation of the circle circum

By sin A+ ya sin B+ aß sin C=0.

125. The geometrical interpretation of the equation just found deserves attention. If from any point O we let fall perpendiculars OP, OQ, on the lines a, ß, then (Art. 54) a, ß are the lengths of these perpendiculars; and since the angle between them is the supplement of C, the

quantity aß sin Cis double the area of the triangle OPQ. In like manner, ya sin B and By sin A are double the triangles OPR, OQR. Hence the quantity

By sin A+ ya sin B+ aß sin C

P

is double the area of the triangle PQR, and the equation found in the last article asserts that if the point O be taken on the circumference of the circumscribing circle, the area PQR will vanish, that is to say (Art. 36, Cor. 2), the three points P, Q, R will lie on one right line.

A

R

B

If it were required to find the locus of a point from which, if we let fall perpendiculars on the sides of a triangle, and join their feet, the triangle PQR so formed should have a constant magnitude, the equation of the locus would be

By sin A+ ya sin B+ aß sin C= constant,

and, since this only differs from the equation of the circumscribing circle in the constant part, it is (Art. 81) the equation of a circle concentric with the circumscribing circle.*

126. The following inferences may be drawn from the equation By+mya + naß = 0, whether or not l, m, n have the values sin A, sin B, sin C, and therefore lead to theorems true not only of the circle but of any curve of the second degree circumscribing the triangle. Write the equation in the form

y (IB +ma) + naß = 0;

and we saw in Art. 124 that y meets the curve in the two points where it meets the lines a and B; since if we make y = 0 in the equation, it reduces to aẞ=0. Now, for the same reason, the two points in which 18+ ma meets the curve are the two points where it meets the lines a and B. But these two points coincide, since 18+ ma passes through the point aß. Hence the line 18+ma, which meets the curve in two coincident points, is (Art. 83) the tangent at the point aß.

In the case of the circle the tangent is a sin B+ẞ sin A. Now we saw (Art. 64) that a sin A+ ẞ sin B denotes a parallel to the base y drawn through the vertex. Hence (Art. 55) the tangent makes the same angle with one side that the base makes with the other (Euc. III. 32).

* Consider a quadrilateral inscribed in a circle of which a, ẞ, y, d are sides and ɛ a diagonal; then the equation of the circle may be written in either of the forms

where A is the angle in the segment subtended by a, &c., and we have written & with

a negative side in the second equation, because opposite sides of the line are considered in the two triangles.

sin D
δ

Hence, every point on the circle satisfies also the equation = 0.

This equation when cleared of fractions is of the third degree, and represents, together with the circle, the line joining the intersections of ay, Bd. In the same manner, if we have an inscribed polygon of any number of sides, Dr. Casey has shewn that an equation of similar form will be satisfied for any point of the circle.