Analytic geometry. Textbook

belong to the circle with the center at the origin O and radius R, the value is the angle between the positive direction of the x-axis and the segment OM, this angle is counted counterclockwise, and coordinates of every point of this circle can be written in the form given above (see Fig. 20).

y

M(R cos ;R sin )

R

O x

Fig. 20.

Now we return to the line with the equation x cos + y sin p = 0. Assume rst that p > 0.

y

O

Fig. 21. Parameters of the the normal equation, p 6= 0

The line does not pass through the origin O, the point M(p cos ; p sin ) belongs to the line. To prove this we substitute the coordinates of this point into the equation:

p cos cos + p sin sin p = p(cos2 + sin2 ) p = 0:

the sign. We present a di erent approach for determining the position of a point on the plane.

Let us take a point O on the plane and a ray OA from this point. The point O (analogous to the origin of a rectangular coordinate system) is called the pole, and the ray is called the polar axis. As in the case of rectangular coordinates, it is assumed that on this plane there is used some unit in which the distances are measured.

Let M be a point on the plane. Its distance jOMj from the pole is called the polar radius and is usually denoted by r or . If M 6= O then the angle

\MOP is called the polar angle. This angle is found taking into account the sign (the counterclockwise direction from the OA to OM is considered positive, the clockwise direction is considered negative) and is de ned up to a term that is a multiple of 2 . In the case M = O the polar angle is considered undetermined.

M(r; ')

r

'

O A

Fig. 23. Polar coordinates

The polar angle is usually denoted by ' or . The values r, ' are called the polar coordinates of the given point. As in the case of rectangular coordinates, the polar coordinates are written in the similar form (r; ').

We will nd the relation between rectangular and polar coordinates. Let the pole and the polar axis of the system of polar coordinates be at the same time the origin and the x-axis of the system of rectangular coor-

dinates respectively. Let M be any point of the plane, (x; y) its rectangular coordinates, and (r; ') its polar coordinates.

Fig. 24. Relations between rectangular and polar coordinates

Then, from the de nition of the trigonometric functions we obtain that

1.4Some tasks on straight lines on the plane

1)Find equation of the line passing through two di erent points

M1(x1; y1) and M2(x2; y2).

We take the pencil of the lines with the center at M1. The lines from this pencil may be de ned by the equations

We need to choose from this set the line passing through the point M2. To do this, we substitute the coordinates of the point M2 into the equation (14):

A(x2 x1) + B(y2 y1) = 0:

This equality is valid if A = y2 y1, B = (x2 x1). We note that at least one of these numbers is nonzero, since otherwise we would get that

M2 = M1. Substituting these values into the equation (14), we obtain the required equation of the line

(y2 y1)(x x1) (x2 x1)(y y1) = 0;

or in the following form that is easier to remember

Using determinants, it is also possible to rewrite this equation in one of the

2) Find the equation of the line parallel to the line Ax + By + C = 0 and passing through the point M(x0; y0).

We will look for the equation of this line in the form

Ax + By + D = 0

where D is the number to be found. Substituting into this equation the coordinates if the point M0 we get that Ax0 + By0 + D = 0, therefore

D = (Ax0 + By0) and the equation of the line takes the form

Ax + By (Ax0 + By0) = 0

or

A(x x0) + B(y y0) = 0:

In other words it is the line from the pencil with the center M0 with coe - cients taken from the given line.

3) Find equation of the line that is orthogonal to the line Ax+By+C = 0 and passes through the point M0(x0; y0).

The line with the equation Bx Ay = 0 is orthogonal to the given line since AB + B( A) = 0. Therefore we are to nd the straight line which is parallel to the line with the equation Bx Ay = 0 and passes through the given point. Using the result of the previous example we get that this line may be de ned by the equation B(x x0) A(y y0) = 0.

Remark. If r = 0 we obtain the equation (x a)2 +(y b)2 = 0 which is satis ed by no values except x = a and y = b. Therefore it de nes the single point (a; b). In this case, it is said that this equation de nes the degenerate circle.

The equation (1) may be transformed as follows

where c stands for a2 + b2 r2.

If the center of the circle is the origin, then a = 0, b = 0, and the equation takes the form x2 + y2 = r2.

Not every equation of the form (2) (with a value c independent of a and b) de nes a real circle. For example, there are no points with the coordinates satisfying the equation x2 + y2 = 1, since the sum x2 + y2,

being the sum of squares, cannot be negative.

Theorem 1. The equation of the form x2 + y2 2ax 2by + c = 0

If a2 + b2 c < 0 there are no points satisfying this equation.

p

Assuming that a2 + b2 c > 0 and letting r = a2 + b2 c we have (x a)2 + (y b)2 = r2:

We get the equation (1) representing the circle with the center (a; b) and radius r.

Remark. When a2 +b2 c is negative, say equals s for some positive s, the equation (3) takes the form

(x a)2 + (y b)2 = s:

There are no points satisfying this equation. In this case, it is said that this equation de nes the imaginary circle. In our case, this de nition is used only to facilitate the formulation of the theorem given below. With this additional agreement, the previous theorem may be reformulated as follows. The equation of the form x2 + y2 2ax 2by + c = 0 represents a circle. This circle is degenerate if and only if c = a2 + b2 and imaginary if and only if c > a2 + b2.

Definition. Equation of the form

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

where A, B and C do not vanish simultaneously if called the second degree equation in x and y.

Theorem 2. A second degree equation in x and y represents a circle

(real, degenerate, or imaginary) if and only if it has no terms in xy and coe cients of x2 and y2 are equal.

Proof. Let us consider the second degree equation

Ax2 + Ay2 + Bx + Cy + D = 0:

Then A 6= 0 since otherwise the coe cients of the second degree terms x2, xy and y2 vanish simultaneously. Dividing the both sides of the equation by A we rewrite it in the form

x2 + y2 + BAx + CAy + DA = 0

which as was shown above is the equation of the circle.

The validity of the converse statement is obvious.

Theorem 3. An equation of the tangent to the circle x2 + y2 = r2 at a point M0(x0; y0) of this circle may be written in the form x0x + y0y = r2.

Proof. We have the circle with the center at the origin O(0; 0). From the school math course we know that the tangent to the circle at the point M0 is orthogonal (perpendicular) to the radius OM0.

y

M0

Fig. 1. Tangent to the circle

!

Therefore OM0 is the normal vector of the tangent. Taking into account

!

the equality OM0 = fx0; y0g and the equation of the straight line passing through the given point and having the given normal vector (page 17, equation (5)), we get the equation of the tangent:

x0(x x0) + y0(y y0) = 0;

or x0x + y0y = x20 + y02. Since the point (x0; y0) is on the circle, we have x20 + y02 = r2 and the equation of the tangent takes the form

x0x + y0y = r2:

Now we generalize the previous theorem.

Theorem 4. Equation of the tangent to the circle

x2 + y2 2ax 2by + c = 0