is

C

sin C'

and centre (0, cot C). The centre will therefore be above, on, or below the

base, according as C is acute, right, or obtuse.

Ex. 2. To solve the last example, 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

y - y' = m (x − x'),

then the equation of the other side, making with this the angle C, will be (Art. 42)

(1 + m tan C) (y − y′′) = (m − tan C) (x − x”).

Eliminating m, the equation of the locus is

tan C{(y-y) (y-y') + (x − x') (x − x'') } + x (y' — y′) − y (x' — x") + x'y" — y'x" =0, which reduces to the equation of the last example if y' = y′′ = 0; x' =+ c, x" =— c. 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. 3. Given base and vertical angle, to 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 — tan C) (y − y) + (1 + m tan C) (x − x') = 0.

Eliminating m, the equation of the locus is

tan C {(y - y') (y − y′′) + (x − x') (x − x′′)} = x (y' — y') − y (x' — x') + x'y" — y'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. 4. Given base and ratio of sides of a triangle, find locus of vertex.

With the same axes as in Ex. 1, if ratio be m : n, we find, for equation of locus,

Hence the locus is a circle, whose centre is on the axis of x, at a distance from origin m2 + n2

c; and which meets the base at the points

m-n

c, and x =

C.

m + n

Since the co-ordinates of the extremities of the base are x=+c, these (Art. 7) are the two points where the base is cut in the ratio m : n.

Ex. 5. Given base of a triangle, and m times square of one side,+ n times square of other; find the locus of vertex.

Ex. 6. Find the locus of a point the square of whose distance from a given point is proportional to its distance from a given right line.

Ex. 7. A line of constant length moves between two fixed right lines, and perpendiculars to the lines are raised at its extremities; find the locus of their intersection.

Ex. 8. In general, 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 p. 48) such that Σ (mr2) may

be constant.

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. 48 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 (my2) = 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. 48, 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

R2 (m) = (mr2) – Σ(mp2),

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

95. We shall next give one or two examples involving the problem of Art. 80, 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 cosa + y sin a − 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. 80) found from the equation

Now, if the roots of this equation be x and x", the x of the middle point of the chord x' + x" 2

will (Art. 7) be -, or, from the theory of equations, will = p cosa. In like manner,

y

the y of the middle point will equal p sin a. Hence the equation of the locus is

=tana,

that is, a right line drawn through the centre perpendicular to the system of parallel chords; since a is the angle made with the axis of x by a perpendicular to the chord

Ex. 2. To find the condition that the intercept made by the circle on the line

x cos a + y sin a = p

should subtend a right angle at the point x'y'.

We found (Art. 94, Ex. 2) the condition that the lines joining the points x"y", x""y" to xy should be at right angles to each other; viz.:

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

Let x"y", "y" be the points where the line meets the circle, then, by the last example,

x" + x" = 2p cosa, x"x"" = p2 - r2 sin2 a, y" + y"" = 2p sin a, y"y" = p2 - r2 cos2 a.

Putting in these values, the required condition is

x2 + y2 - 2px' cos a 2py' sina + 2p2 — r2 = 0.

Ex. 3. To find the locus of the middle point of a chord which subtends a right angle at a given point.

If x and y be the co-ordinates of the middle point, we have, by Ex. 1,

p cos α = x, p sin a = y, p2 = x2 + y2,

and, substituting these values, the condition found in the last example becomes

(x − x')2 + (y - y')2 + x2 + y2 = r2,

Ex. 4. To find the locus of the foot of a perpendicular from x'y' on a chord which subtends a right angle at that point.

The co-ordinates of the foot of perpendicular are determined by the equations

x cosa + y sin a=p; (x − x' ́) sin a − (y — y') cos a =

whence, if we write for shortness,

= 0;

we have

R sin ay-y', R cosa = x - x', Rp = x2 + y2 — xx' — yy' ;

but the condition in Ex. 2 may be written

0 = x2+ y22―r2+2p(p-x'cos a-y'sin a) = x2+ y2- r2 + 2p {(x − x') cos a +(y−y')sina};

or, the locus is the same as that found in the last example.

Ex. 5. Given a line and a circle, to find a point such that if any chord be drawn through it, and perpendiculars let fall from its extremities on the given line, the rectangle under these perpendiculars may be constant.

Take the given line for axis of y, and let the axis of x be the perpendicular on it from the centre of the given circle, whose length we shall call p. Then the equation of

Again, if the co-ordinates of the sought point be x', y', the equation of any line through it will be

(y − y) = m(x − x'), or y = mx + y' — mx'.

Substitute this value of y in the equation of the circle, and we shall get, to determine the x of the points where the line meets the circle,

(1 + m2) x2 + {2m (y' — mx') — 2p} x + (y' − mx')2 + p2 − r2 = 0.

-

But x is the perpendicular on the given line, and the product of the two perpendiculars (by the theory of equations)

This will not be independent of m, unless the numerator be divisible by 1 + m2, and it will be found that this cannot be the case unless y' = O and x2 = p2 - r2. Hence there are two such points situated on the axis of x, and at a distance from the origin the tangent drawn from it to the given circle.

=

Ex. 6. If any chord be drawn through a fixed point on a diameter of a circle, and its extremities joined to either end of the diameter, the joining lines will cut off on the tangent, at the other end of the diameter, portions whose rectangle is constant.

Let us take the diameter for axis of x, and either extremity of it for origin, then (Art. 77, Cor. 2) the equation of the circle will be x2 + y2 = 2rx, and that of any chord through a fixed point on the diameter will be y = m (x − x'). By combining these equations we can determine the co-ordinates of the extremities of the chord. We can, however, without solving for these co-ordinates, obtain directly from the equations the equation of the lines joining these extremities to the origin. For if, by combining the equations, we can obtain a homogeneous function of the second degree, it will be, by Art. 68, the equation of two right lines drawn through the origin, and it evidently must be satisfied by the co-ordinates of the points which satisfy the two given equations.

Write these equations thus,

x2 + y2 = 2rx, and mx' = mx − y,

and, multiplying them together, we get

mx' (x2 + y2) = 2rx (mx − y).

This being homogeneous in x and y, is the required equation of the joining lines. It may be written thus,

This equation enables us to find the values of y corresponding to any value of r, and

The intercepts made on a perpendicular at the extremity of the diameter are found by

making x = 2r in the preceding equation, and their product is 4r2:

constant as long as x' is constant.

x' - - 2r

96. We shall next obtain one or two of the properties of the polar of a point from its equation (Art. 86).

If any chord be drawn through a fixed point and tangents at its extremities: to find the locus of their intersection.

N

Let any point on the locus be XY, then the chord joining points of contact of tangents passing through XY is

Xx + Yy = r2;

but by hypothesis, this line passes through the point x'y', therefore, Xx′ + Yy' = r2;

this is the relation connecting the co-ordinates of the point XY, its locus, therefore, is the line

xx' + yy' = r2,

or the polar of the point x'y'.

The proposition just proved may be stated otherwise, thus: If one point lie on the polar of a second point, the second point will lie on the polar of the first point.

For the condition that x'y' should lie on the polar of x"y" is

x'x" + y'y" = r2.

But this is also the condition that the point a"y" should lie on the polar of x'y'.

97. Given any point O, and any two lines through it ; join both directly and transversely the points in which these lines meet a circle; then, if the direct lines intersect each other in P and the transverse in Q, the line PQ will be the polar of the point O, with regard to the circle.

Take the two fixed lines for axes, and let the intercepts made on them by the circle be a and a', b and b'.

Then

will be the equations of the direct lines; and

the equations of the transverse lines. Now, the equation of the