will not become indeterminate when the two points are made to coincide. For, since

r2=x"2 + y" = x"2+y", we have x" — x" = y" — y′′,

—

And if we now make x' = x" and y' =y", we find for the equation of the tangent

y-y' ac'

X

==

y'

or, reducing, and remembering that x" + y22 = r2, we get finally xx' + yy' = r*.

Otherwise thus: The equation of the chord joining two points on a circle may be written

(x − x′) (x − x') + (y − y') (y − y′′) = x2 + y2 — r2.

For this is the equation of a right line, since the terms x2+ y2 on each side destroy each other; and if we make x = y=y', the left-hand side vanishes identically, and the right-hand side vanishes, since the point x'y' is on the circle. In like manner the equation is satisfied by the coordinates "y". This then is the equation of a chord; and the equation of the tangent got by making '=x", y'=y", is

(x − x')2 + (y — y')2 = x2 + y3 — r3, which reduced, gives, as before, xx' + yy' = r2.

If we were now to transform the equations to a new origin, so that the coordinates of the centre should become a, B, we must substitute (Art. 8) xa, x'-a, y-B, y' - B, for x, x', y, y', respectively; the equation of the circle would become

a form easily remembered from its similarity to the equation of the circle.

This method is due to Mr. Burnside.

COR. The tangent is perpendicular to the radius, for the equation of the radius, the centre being origin, is easily seen to be x'y-y'x=0; but this (Art. 32) is perpendicular to xx' + yy' = r2.

86. The method used in the last article may be applied to the general equation*

ax2+2hxy + by* + 2gx+2fy+c=0.

The equation of the chord joining two points on the curve may be written

=

a (x − x′) (x − x′′) + 2h (x − x') (y − y′′) + b (y − y') (y − y′′) = ax2 + 2hxy + by* + 2gx + 2fy + c. For the equation represents a right line, the terms above the first degree destroying each other; and, as before, it is evidently satisfied by the two points on the curve x'y', "y". Putting x"=x', y′′ =y', we get the equation of the tangent a(x-x')2+2h(x-x')(y-y')+b(y-y')'=ax2+2hxy+by+2gx+2fy+c;

or, expanding,

2ax'x + 2h (x'y + y'x) + 2by'y + 2gx+2fy + c = ax"*+2hx'y' + by”. Add to both sides 2gx' + 2fy' + c, and the right-hand side will vanish, because x'y' satisfies the equation of the curve. Thus the equation of the tangent becomes

ax'x + h (x'y + y'x) + by'y + g (x + x') + ƒ (y + y') + c = 0.

This equation will be more easily remembered if we compare it with the equation of the curve, when we see that it is derived from it by writing x'x and y'y for x and y2, x'y+y'x for 2xy, and x' + x, y' + y for 2x and 2y.

Ex. 1. Find the equations of the tangents to the curves xy = c2 and y2 = px.
Ans. x'y+y'x2c2 and 2yy' = p (x + x).

Ex. 2. Find the tangent at the point (5, 4) to (x − 2)2 + (y − 3)2 = 10.

Ans. 3x+y= 19.

Ex. 3. What is the equation of the chord joining the points x'y', x"y" on the circle x2+ y2= p2? Ans. (x' + x') x + (y′ + y′′) y = r2 + x'x" + y'y".

Ex. 4. Find the condition that Ax + By + C = 0 should touch

Ans.

Aa + BB + C
=r; since the perpendicular on the line from aß is equal to r.
J(A2 + B2)

Of course when this equation represents a circle we must have b=a, ha cos w; but since the process is the same, whether or not bor h have these particular values, we prefer in this and one or two similar cases to obtain at once formula which will afterwards be required in our discussion of the general equation of the second degree.

M

87. To draw a tangent to the circle x2+ y2= r2 from any point x'y'. Let the point of contact be a"y", then since, by hypothesis, the coordinates x'y' satisfy the equation of the tangent at "y", we have the condition x'x" + y'y" = r2.

And since "y" is on the circle, we have also

These two conditions are sufficient to determine the coordinates x", y". Solving the equations we get

12

Hence, from every point may be drawn two tangents to a circle. These tangents will be real when x” + y” is > r*, or the point outside the circle; they will be imaginary when "+y" is <r”, or the point inside the circle; and they will coincide when x2+y" = r2, or the point on the circle.

88. We have seen that the coordinates of the points of contact are found by solving for x and y from the equations xx' + yy' = r2; x2 + y* = r2.

Now the geometrical meaning of these equations evidently is, that these points are the intersections of the circle x2+ y2 = r2 with the right line xx' + yy' = r2. This, last, then is the equation of the right line joining the points of contact of tangents from the point x'y'; as may also be verified by forming the equation of the line joining the two points whose coordinates were found in the last article.*

We see, then, that whether the tangents from x'y' be real or imaginary, the line joining their points of contact will be the real line xx+yy=r", which we shall call the polar of x'y' with regard to the circle. This line is evidently perpendicular to the

* In general the equation of the tangent to any curve expresses a relation connecting the coordinates of any point on the tangent, with the coordinates of the point of contact. If we are given a point on the tangent and required to find the point of contact, we have only to accentuate the coordinates of the point which is supposed to be known, and remove the accents from those of the point of contact, when we have the equation of a curve on which that point must lie, and whose intersection with the given curve determines the point of contact. Thus, if the equation of the tangent to a curve at any point x'y' be xx" + yy"=3, the points of contact of tangents drawn from any point x'y' must lie on the curve x'x2+y'y2 = 1+3. It is only in the case of curves of the second degree that the equation which determines the points of contact is similar in form to the equation of the tangent.

line (x'y-y'x=0), which joins x'y' to the centre; and its dis

tance from the centre (Art. 23) is

Hence, the polar of any point P is constructed geometrically by joining it to the centre C, taking on the joining line a point M, such that CM.CP= r2, and erecting a perpendicular to CP at M. We see, also, that the equation of the polar is similar in form to that of the tangent, only that in the former case the point x'y' is not supposed to be necessarily on the circle; if, however, x'y' be on the circle, then its polar is the tangent at that point.

89. To find the equation of the polar of x'y' with regard to the ax2 + 2hxy + by2 + 2gx + 2ƒy + c = 0.

curve

We have seen (Art. 86) that the equation of the tangent is

ax'x + h (x'y+y'x) + by'y + g (x + x') + f (y + y') + c = 0. This expresses a relation between the coordinates xy of any point on the tangent, and those of the point of contact x'y'. We indicate that the former coordinates are known and the latter unknown, by accentuating the former, and removing the accents from the latter coordinates. But the equation, being symmetrical with respect to the coordinates xy, x'y', is unchanged by this operation. The equation then written above (which when x'y' is a point on the curve, represents the tangent at that point), when x'y' is not on the curve, represents a line on which lie the points of contact of tangents real or imaginary from x'y'.

If we substitute x'y' for xy in the equation of the polar we get the same result as if we made the same substitution in the equation of the curve. This result then vanishes when x'y' is on the curve. Hence the polar of a point passes through that point only when the point is on the curve, in which case the polar is the tangent.

COR. The polar of the origin is gx+fy+c=0.

Ex. 1. Find the polar of (4, 4) with regard to (x-1)2+(y-2)2=13. Ans. 3x+2y=20. Ex. 2. Find the polar of (4, 5) with regard to x2+y2-3x-4y=8. Ans. 5x+6y=48, Ex. 3. Find the pole of Ax + By + C = 0 with regard to x2 + y2 = p2.

Ar2

Ans.

(- 4m2, – Br3),

as appears from comparing the given equation with
xx' + yy' = r2.

Ex. 4. Find the pole of 3x + 4y = 7 with regard to x2 + y2 = 14. Ans. (6,8).
Ex. 5. Find the pole of 2x + 3y = 6 with regard to (x − 1)2 + (y − 2)2 = 12.
Ans. (-11, 16).

90. To find the length of the tangent drawn from any point to the circle (x − a)2 + (y − B)2 — y2 = 0.

The square of the distance of any point from the centre

= (x − a)2+(y-B)*;

and since this square exceeds the square of the tangent by the square of the radius, the square of the tangent from any point is found by substituting the coordinates of that point for x and y in the first member of the equation of the circle

when divided by a, is (Art. 80) equivalent to one of the form (x − a)2+(y-ẞ)2 — r2 = 0,

we learn that the square of the tangent to a circle whose equation is given in its most general form is found by dividing by the coefficient of x2, and then substituting in the equation the coordinates of the given point.

The square of the tangent from the origin is found by making x and y = 0, and is, therefore, the absolute term in the equation of the circle, divided by a.

The same reasoning is applicable if the axes be oblique.

#91. To find the ratio in which the line joining two given points x'y', x"y", is cut by a given circle.

We proceed precisely as in Art. 42. The coordinates of point on the line must (Art. 7) be of the form

any

and arranging, we have, to determine the ratio 7: m, the quadratic

l2 (x''2 + y'12 — y2) + 2lm (x'x" + y'y" — r3) + m2 (x2+y” — r2) = 0. · The values of 1:m being determined from this equation, we have at once the coordinates of the points where the right line meets the circle. The symmetry of the equation makes this method sometimes more convenient than that used (Art. 82).