chord will (Art. 7) be (x' + x"), or, from the theory of equations, will = p cos a. In like manner, the y of the middle point will equal p sin a. Hence the equation of the locus is y =x 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 any of the chords.

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. 96, 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.

Let x"y", x""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

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 a = 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. 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 x, and let the axis of y be the perpendicular on it from the centre of the given circle, whose equation will then be

x2 + (y − ẞ)2 = p2.

Let the co-ordinates of the sought point be x'y'; then the equation of any line through it will be y-y' = '= m (x − x'). Eliminate x between these two equations and we get a quadratic for y, the product of whose roots will be found to be

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 x = 0, y'2 = B2 — r2.

Ex. 5. To find the condition that the intercept made on x cosa + y sin a — -Pr by the circle

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

may subtend a right angle at the origin. The equation of the pair of lines joining the extremities of the chord to the origin may be written down at once. For if we multiply the terms of the second degree in the equation of the circle by p2, those of the first degree by p (x cos a + y sin a), and the absolute term by (x cos a + y sin a)2, we get an equation homogeneous in x and y, which therefore represents right lines drawn through the origin; and it is satisfied by those points on the circle for which x cos a + y sin a=p. The equation expanded and arranged is

(p2 + 2gp cos a + c cos2 a) x2 + 2 (g sin a +f cos a + c sin a cosa) xy

+ (p2 + 2fp sin a + c sin2a) y2 = 0.

These two lines cut at right angles (Art. 74) if

2p2+2p (g cos a +f sin a) + c = 0.

Ex. 6. To find the locus of the foot of the perpendicular from the origin on a chord which subtends a right angle at the origin. The polar co-ordinates of the locus are p and a in the equation last found; and the equation of the locus is therefore

2(x2 + y2) + 2gx + 2fy + c = 0.

It will be found on examination that this is the same circle as in Ex. 3.

Ex. 7. 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 cut off on the tangent at the other end, portions whose rectangle is constant.

Find, as in Ex. 5, the equation of the lines joining to the origin the intersections of x2 + y2 – 2rx with the chord y = m (x − x') which passes through the fixed point (x', 0). The intercepts on the tangent are found by putting x = 2r in this equation and seeking the corresponding values of y. The product of these values will be

98. We shall next obtain from the equations (Art. 88) a few of the properties of poles and polars.

If a point A lie on the polar of B, then B lies on the polar of A. For the condition that x'y' should lie on the polar of "y" is x'x"+y'y" = r2; but this is also the condition that the point x"y" should lie on the polar of x'y'. It is equally true if we use the general equation (Art. 89) that the result of substituting the co-ordinates "y" in the equation of the polar of x'y' is the same as that of substituting the co-ordinates x'y' in the polar of x"y". This theorem then, and those which follow, are true of all curves of the second degree. It may be otherwise stated thus: if the polar of B pass through a fixed point A, the locus of B is the polar of A.

99. Given a circle and a triangle ABC, if we take the polars with respect to the circle, of A, B, C'; we form a new triangle A'B'C' called the conjugate triangle, A' being the pole of BC, B' of CA, and C' of AB. In the particular case where the polars of A, B, C respectively are BC, CA, AB, the second triangle coincides with the first, and the triangle is called a self-conjugate triangle.

The lines AA', BB', CC', joining the corresponding vertices of a triangle and of its conjugate, meet in a point.

The equation of the line joining the point x'y' to the inter

section of the two lines xx" + yy" — r2 = 0 and xx" + yy"" − r2 = 0 is (Art. 40, Ex. 3)

AA', (x'x'"+y'y"" — r*) (xx" + yy" — 12)

In like manner

BB', (x'x"+y'y" — x2) (xx'" + yy"" — p2)

— (x"x" + y′′y" — r2) (xx' + yy' — r2) = 0 ;

and CC, ("x""'+y"y" — r2) (xx' + yy' — r2).

— (x'x''' + y'y'” — 72) (xx" + yy′′ — r2) = 0 ; and by Art. 41 these lines must pass through the same point. The following is a particular case of the theorem just proved : If a circle be inscribed in a triangle, and each vertex of the triangle joined to the point of contact of the circle with the opposite side, the three joining lines will meet in a point.

The proof just given applies equally if we use the general equation. If we write for shortness P=0 for the equation of the polar of x'y', (ax'x + &c. = 0); and in like manner P2, P for the polars of "y", ""y""; and if we write [1, 2] for the result of substituting the co-ordinates "y" in the polar of x'y', (ax'x"+&c.), then the equations are easily seen to be

which denote three lines meeting in a point. It follows (Art. 60, Ex. 3) that the intersections of corresponding sides of a triangle and its conjugate lie in one right line.

100. 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 λ and λ', μ and μ'. Then

will be the equations of the direct lines; and

for (see Art. 40) this line passes through the intersection of

λ and

ax2 + 2hxy + by2+2gx + 2fy + c = 0,

are determined from the equation ax* + 2gx+c=0

but we saw (Art. 89) that this was the equation of the polar of the origin 0. Hence it appears that if the point ◊ were given, and the two lines through it were not fixed, the locus of the points P and Q would be the polar of the point 0.

101. Given any two points A and B, and their polars with respect to a circle whose centre is 0: let fall a perpendicular AP from A on the polar of B, and a perpendicular BQ from B on the ОА polar of A; then

OB

=
AP BQ'

The equation of the polar of A (x'y') is xx′ + yy' — r2=0; and BQ, the perpendicular on this line from B(x"y"), is (Art. 34)

Hence, since √(x22 + y'2) =OA, we find

OA.BQ=x'x" + y'y" — r2 ;

102. In working out questions on the circle it is often convenient, instead of denoting the position of a point on the curve by its two co-ordinates x'y', to express both these in terms of a single independent variable. Thus, let ' be the angle which the radius to x'y' makes with the axis of x, then x'=r cose', y'r sine', and on substituting these values our formule will generally become simplified.

The equation of the tangent at the point x'y' will by this substitution become

x cose'y sine'=r;

and the equation of the chord joining x'y', x"y", which (Art. 86, Ex. 3) is

x (x' + x') + y (y' + y′′) = r2 + x'x′′ +y'y",

will, by a similar substitution, become

x cos § (0′ + 0′′) + y sin § (0′ + 0′′) = r cos § (0′ — 0′′),

✔' and " being the angles which radii drawn to the extremities of the chord make with the axis of x.

This equation might also have been obtained directly from the general equation of a right line (Art. 23) x cosa + y sina=Pr for the angle which the perpendicular on the chord makes with the axis is plainly half the sum of the angles made with the axis by radii to its extremities; and the perpendicular on the chord

=r cos1⁄2 (0′ — 0′′).

Ex. 1. To find the co-ordinates of the intersection of tangents at two given points on the circle. The tangents being

x cos 0' + y sin 0′ = r, x cos 0" + y sin 0” = r,

the co-ordinates of their intersection are

x=r

cos (0'+0")
, y = r
cos (0'- 0")'

sin (0'+0")

cos (0'- 0")

Ex. 2. To find the locus of the intersection of tangents at the extremities of a chord whose length is constant.

it reduces to cos (0′ – 0") = constant, or 'e" constant. If the given length of