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If "y" lie on the polar of x'y', we have x'x" +y'y" — r2 = 0 (Art. 88), and the factors of the preceding equation must be of the form l+μm, l-μm; the line joining x'y', x"y" is therefore cut internally and externally in the same ratio, and we deduce the well-known theorem, any line drawn through a point is cut harmonically by the point, the circle, and the polar of the point.

*92. To find the equation of the tangents from a given point to a given circle.

We have already (Art. 87) found the co-ordinates of the points of contact; substituting, therefore, these values in the equation xx" + yy" — "=0, we have for the equation of one tangent

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r (xx' + yy' — x22 — y') + (xy' — yx') √(x'2 + y'2 — r2) = 0, and for that of the other,

r (xx' + yy' — x” — y′′) -- (xy' — yx') √ (x'2 + y22 − r2) = 0. These two equations multiplied together give the equation of the pair of tangents in a form free from radicals. The preceding article enables us, however, to obtain this equation in a still more simple form. For the equation which determines 7: m will have equal roots if the line joining x'y', x"y" touch the given circle; if then "y" be any point on either of the tangents through x'y', its co-ordinates must satisfy the condition

(x12 + y'2 — r2) (x2 + y2 — r2) = (xx′ + yy' — r2)2.

This, therefore, is the equation of the pair of tangents through the point x'y'. It is not difficult to prove that this equation is identical with that obtained by the method first indicated.

The process used in this and the preceding article is equally applicable to the general equation. We find in precisely the same way that 7: m is determined from the quadratic

l2 (ax”2 + 2hx'"'y" + by"2 + 2gx" +2fy" +c)

+ 2lm {ax'x" + h (x'y" + x'y')+by'y" + g(x' +x") +ƒ (y′+y′′)+c} + m2 (ax2+2hx'y' + by" + 2gx' + 2ƒy' + c) = 0 ; from which we infer, as before, that when "y" lies on the polar of x'y' the line joining these points is cut harmonically; and also that the equation of the pair of tangents from x'y' is

(ax” +2hx'y' + by'" + 2gx' + 2fy'+c) (ax2+2hxy+by2+2gx+2fy+c) = {ax'x + h (x'y + xy') + byy' +9 (x+x') +ƒ (y + y') + c}2.

93. To find the equation of a circle passing through three

given points.

We have only to write down the general equation

x2 + y2+2gx+2fy + c = 0,

and then substituting in it, successively, the co-ordinates of each of the given points, we have three equations to determine the three unknown quantities g, f, c. We might also obtain the equation by determining the co-ordinates of the centre and the radius, as in Ex. 5, p. 4.

Ex. 1. Find the circle through (2, 3), (4, 5), (6, 1).

Ans. (x)2 + (y - §)2 = 50 (see p. 4).

Ex. 2. Find the circle through the origin and through (2, 3) and (3, 4). Here c = 0, and we have 13+ 4g +6ƒ = 0, 25+ 6g+8f=0, whence 2g=-23, 2ƒ= 11. Ex. 3. Taking the same axes as in Art. 48, Ex. 1, find the equation of the circle through the origin and through the middle points of sides; and show that it also passes through the middle point of base.

Ans. 2p (x2 + y2) − p (s — s'′) x − ( p2 + ss′) y = 0.

*94. To express the equation of the circle through three points x'y', x"y", x"y" in terms of the co-ordinates of those points. We have to substitute in

x2+ y2+2gx+2fy+c=0,

the values of g, f, c derived from

(x22 + y22)+2gx' +2fy' +c=0,

(x2 + y2)+2gx" + 2fy" +c=0,

(x'''2 + y'''2)+2gx"" +2ƒy"" +c=0.

The result of thus eliminating g, f, c between these four equations will be found to be

(x12 +y") {x" (y"

+(x2 + y2) {x''' ( y

- (x'''2 + y'''2) {x (y'

−y′′)+x′ (y" − y )+x" (y — y′ )} = 0,

as may be seen by multiplying each of the four equations by the quantities which multiply (x + y) &c. in the last written equation, and adding them together, when the quantities multiplying 9, f, c will be found to vanish identically.

If it were required to find the condition that four points should lie on a circle, we have only to write x, y, for x and y

in the last equation. It is easy to see that the following is the geometrical interpretation of the resulting condition. If A, B, C, D be any four points on a circle, and O any fifth point taken arbitrarily, and if we denote by BCD the area of the triangle BCD, &c., then

OA.BCD+OC2.ABD=0B2. A CD + OD2. ABC.

95. We shall conclude this chapter by showing how to find the polar equation of a circle.

We may either obtain it by substituting for x, p cose, and for y, p sine (Art. 12), in either of the equations of the circle already given,

a (x2 + y2) + 2gx + 2fy+c=0, or (x − a)2 + ( y − B)2 = r2, or else we may find it independently, from the definition of the circle, as follows:

Let O be the pole, C the centre of the circle, and OC the fixed axis; let the distance OC=d,

and let OP be any radius vector, and, therefore, =p, and the angle POC=0, then we have

PC2=OP"+0C-20P.OC cos POC,

that is,

or

p2 - 2dp coз0+ d2 — r" = 0.

This, therefore, is the polar equation of the circle.

P

Q

If the fixed axis did not coincide with OC, but made with it any angle a, the equation would be, as in Art. 44,

If we suppose the pole on the circle, the equation will take a simpler form, for then r=d, and the equation will be reduced to p=2r cos 0,

a result which we might have also obtained at once geometrically from the property that the angle in a semicircle is right; or else by substituting for x and y their polar values in the equation (Art. 79) x2+ y2 = 2rx.

CHAPTER VII.

THEOREMS AND EXAMPLES ON THE CIRCLE.

96. HAVING in the last chapter shown how to form the equations of the circle, and of the most remarkable lines related to it, we proceed in this chapter to illustrate these equations by examples, and to apply them to the establishment of some of the principal properties of the circle. We recommend the

reader first to refer to the answers to the examples of Art. 49, to examine in each case whether the equation represents a circle, and if so to determine its position either (Art. 80) by finding the co-ordinates of the centre and the radius, or (Art. 84) by finding the points where the circle meets the axes. We add a few more examples of circular loci.

Ex. 1. Given base and vertical angle, find the locus of vertex, the axes having any position.

Let the co-ordinates of the extremities of base be x'y', x"y". Let the equation of one side be

then the equation of the other side, making with this the angle C, will be (Art. 33) (1+m tanC) (y — y′′) = (m − tan C′) (x − x′).

Eliminating m, the equation of the locus is

tanC {(y — y') (y — y'') + (x − x′) (x − x'')} + x (y' — y′) — y (x′ — x'') + x'y′′ — y′x" = 0.

If C be a right angle, the equations of the sides are

y-y' ' = m (x − x'); m (y − 'y'') + (x − x') = 0,

and that of the locus

(y - y') (y − y') + (x − x′) (x − x′′) = 0.

Ex. 2. Given base and vertical angle, find the locus of the intersection of perpendiculars of the triangle.

The equations of the perpendiculars to the sides are

m (y − y′′) + (x − x′′) = 0, (m − tanC) (y − y') + (1 + m tanC) (x − x') = 0.

Eliminating m, the equation of the locus is

= x (y' — y′′) — y (x' — x') + x'y" — y'x" ;

tan C {(y-y') (y − y') + (x − x′) (x − x')} an equation which only differs from that of the last article by the sign of tan C, and which is therefore the locus we should have found for the vertex had we been given the same base and a vertical angle equal to the supplement of the given one.

Ex. 3. Given any number of points, to find locus of a point such that m' times square of its distance from the first +m" times square of its distance from the second + &c. = a constant; or (adopting the notation used in Ex 4, p. 49) such that Σ (mr2) may be constant.

The square of the distance of any point xy from x'y' is (x − x′)2 + (y − y')2. Multiply this by m', and add it to the corresponding terms found by expressing the distance of the point xy from the other points x"y", &c. If we adopt the notation of p. 49, we may write for the equation of the locus,

Σ (m) x2 + Σ (m) y2 – 2Σ (mx') x − 2Σ (my') y + Σ (mx'2) + Σ (my'2) = C. Hence the locus will be a circle, the co-ordinates of whose centre will be

that is to say, the centre will be the point which, in p. 50, was called the centre of mean position of the given points.

If we investigate the value of the radius of this circle, we shall find

where (mr)2=C= sum of m times square of distance of each of the given points from any point on the circle, and Σ (mp2) sum of m times square of distance of each point from the centre of mean position.

Ex. 4. Find the locus of a point 0, such that if parallels be drawn through it to the three sides of a triangle, meeting them in points B, C; C', A'; A", B"; the sum may be given of the three rectangles

BO.OC+C'0.0A' + A”0.0B".

Taking two sides for axes, the equation of the locus is

This represents a circle, which, as is easily seen, is concentric with the circumscribing circle; the co-ordinates of the centre in both cases being given by the equations 2 (a + ẞ cos C') = a, 2 (ẞ + a cos C) = b. These last two equations enable us to solve the problem to find the locus of the centre of circumscribing circle, when two sides of a triangle are given in position, and any relation connecting their lengths is given.

Ex. 5. Find the locus of a point O, if the line joining it to a fixed point makes the same intercept on the axis of x, as is made on the axis of y by a perpendicular through O to the joining line.

Ex. 6. Find the locus of a point, such that if it be joined to the vertices of a triangle, and perpendiculars to the joining lines erected at the vertices, these perpendiculars meet in a point.

97. We shall next give one or two examples involving the problem of Art. 82, to find the co-ordinates of the points where a given line meets a given circle.

Ex. 1. To find the locus of the middle points of chords of a given circle, drawn parallel to a given line.

Let the equation of any of the parallel chords be

x cos ay sina - p = 0,

where a is, by hypothesis, given, and p is indeterminate; the abscissæ of the points where this line meets the circle are (Art. 82) found from the equation

Now, if the roots of this equation be x' and x", the x of the middle point of the

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