Chapter 9 Objects in the Two-Dimensional Coordinate System

Section 9.3 Circles in the Plane

9.3.4 Relative Positions of Circles

We may also raise the question of the relative position in the coordinate system for a circle and a line or for two circles, as we did for two lines. That is, we have to determine whether the two objects intersect, osculate (touch in a single point), or do not have any points in common. In the case of a circle and a line, the intersection point or the osculation point can be calculated easily. For two circles, this is more difficult and goes beyond the scope of this course. For two circles we will only discuss whether or not they intersect or osculate, not at which points this happens.

Info 9.3.9

Let a circle

K

and a line

g

in the plane be given (by equations in coordinate form with respect to a fixed coordinate system). Then the circle and the line have exactly one of the following three relative positions with respect to each other.

The circle $K$ and the line $g$ do not have any points in common. This is true for the red line in the figure below. Such a line is called an exterior line to the circle.
The circle $K$ and the line $g$ have exactly one point in common, i.e. the line osculates the circle. This is true for the blue line in the figure below. Such a line is called a tangent line to the circle.
The circle $K$ and the line $g$ have two points in common, i.e. the line intersects the circle. This is true for the green line in the figure below. Such a line is called a secant line to the circle.

If a line is a tangent or a secant, the circle and the line have one or two points in common, respectively. How can the intersection point or points be calculated? Since the points lie on both the circle and on the line, they have to satisfy both the equation of the circle and the equation of the line in coordinate form. Thus, we have two equations for the two unknown coordinates of the intersection points, and we are able to calculate the coordinates. However, since in the equations of a circle in coordinate form the unknown coordinates are squared, we do not have two linear equations. Hence, the methods for solving systems of linear equations described in Module 5 unfortunately cannot be applied. The Info Box below outlines the method for calculating the intersection points.

Info 9.3.10

Let a circle

K

in the plane with centre

(x_{0}; y_{0})

and radius

r

be given by an equation of a circle in coordinate form

K : (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2},

and a line with slope

m

and

y

-intercept

b

by an equation in normal form

g : y = m x + b .

To calculate the potential intersection points, the equation of

g

can be substituted into the equation of

K

resulting in a quadratic equation in the variable

x

(x - x_{0})^{2} + (\underset{= y}{\underset{⏟}{m x + b}} - y_{0})^{2} = r^{2} .

For a quadratic equation three cases can occur (see Section 2.1.5):

The quadratic equation has no solution. In this case, $g$ is an exterior line to $K$ . An $x$ -coordinate of a common point cannot be found.
The quadratic equation has exactly one solution. In this case, $g$ is a tangent line to $K$ . For the $x$ -coordinate (the solution of the quadratic equation), a corresponding $y$ -coordinate can be calculated from the given equation of a line. The two coordinates define the osculation point of the tangent line $g$ and the circle $K$ .
The quadratic equation has two solutions. In this case, $g$ is a secant line to $K$ . For both $x$ -coordinates (the two solutions of the quadratic equation), corresponding $y$ -coordinates can be calculated from the given equation of a line. The two pairs of coordinates define the intersection points of the secant line and the circle $K$ .

Example 9.3.11

Let a circle

K

with the centre

(2; 2)

and a radius of

1

be given by

K : (x - 2)^{2} + (y - 2)^{2} = 1,

and the lines

g_{1}

g_{2}

, and

g_{3}

g_{1} : y = x - \sqrt{2}

g_{2} : y = x + 1

g_{3} : y = 2 x + 2 .

Line $g_{1}$ :
Substituting the equation of the line $g_{1}$ into the given equation of the circle results in

$(x - 2)^{2} + (x - \sqrt{2} - 2)^{2} = 1 \Leftrightarrow (x - 2)^{2} + [(x - 2) - \sqrt{2}]^{2} = 1$

$\Leftrightarrow (x - 2)^{2} + (x - 2)^{2} - 2 \sqrt{2} (x - 2) + (\sqrt{2})^{2} = 1 \Leftrightarrow 2 (x - 2)^{2} - 2 \sqrt{2} (x - 2) + 1 = 0$

$\Leftrightarrow (\sqrt{2} (x - 2) - 1)^{2} = 0 \Leftrightarrow \sqrt{2} (x - 2) - 1 = 0 \Leftrightarrow x - 2 = \frac{1}{\sqrt{2}} \Leftrightarrow x = 2 + \frac{1}{\sqrt{2}} .$
The resulting quadratic equation has only the solution $x = 2 + \frac{1}{\sqrt{2}}$ . Thus, $g_{1}$ is a tangent line to $K$ that osculates $K$ in a point with the $x$ -coordinate $2 + \frac{1}{\sqrt{2}}$ . The corresponding $y$ -coordinate is calculated from the given equation of the line:

$y = 2 + \frac{1}{\sqrt{2}} - \sqrt{2} = 2 + \frac{\sqrt{2}}{2} - \sqrt{2} = 2 - \frac{1}{\sqrt{2}} .$
Thus, the tangent line $g_{1}$ osculates the circle $K$ at the point $P = (2 + \frac{1}{\sqrt{2}}; 2 - \frac{1}{\sqrt{2}})$ .
Line $g_{2}$ :
Substituting the equation of the line $g_{2}$ into the given equation of the circle results in

$(x - 2)^{2} + (x + 1 - 2)^{2} = 1 \Leftrightarrow (x - 2)^{2} + (x - 1)^{2} = 1$

$\Leftrightarrow x^{2} - 4 x + 4 + x^{2} - 2 x + 1 = 1 \Leftrightarrow 2 x^{2} - 6 x + 4 = 0 \Leftrightarrow x^{2} - 3 x + 2 = 0$

$\Leftrightarrow x_{1,2} = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm 1}{2} \Leftrightarrow x_{1} = 2 \land x_{2} = 1 .$
The resulting quadratic equation has two solutions. Thus, $g_{2}$ is a secant line to the circle $K$ that intersects $K$ in two points with the $x$ -coordinates $x_{1} = 2$ and $x_{2} = 1$ . The corresponding $y$ -coordinates are calculated from the given equation of the line:

$y_{1} = x_{1} + 1 = 2 + 1 = 3 and y_{2} = x_{2} + 1 = 1 + 1 = 2 .$
Thus, the two points $Q_{1} = (2; 3)$ and $Q_{2} = (1; 2)$ are the intersection points of the secant line $g_{2}$ with the circle $K$ .
Line $g_{3}$ :
Substituting the equation of the line $g_{3}$ into the given equation of the circle results in

$(x - 2)^{2} + (2 x + 2 - 2)^{2} = 1 \Leftrightarrow (x - 2)^{2} + (2 x)^{2} = 1 \Leftrightarrow x^{2} - 4 x + 4 + 4 x^{2} = 1$

$\Leftrightarrow 5 x^{2} - 4 x + 3 = 0 .$
The discriminant (see Section 2.1.5) of this quadratic equation is

$(- 4)^{2} - 4 \cdot 5 \cdot 3 = 16 - 60 < 0 .$
Thus, the equation does not have a solution, and the line $g_{3}$ is an exterior line to the circle $K$ .

The circle, all three lines, and the intersection points are shown in the figure below.

Exercise 9.3.12

Let a circle

K

be given by the equation

K : x^{2} + (y - 1)^{2} = 2,

and two lines by

g : y = \sqrt{7} x + 5

h : y = 2 x + 2 .

Show that $g$ is a tangent line to $K$ , and calculate the $x$ -coordinate and $y$ -coordinate of the osculation point $(x; y)$ of $g$ and $K$ .
$x =$

$y =$
Show that $h$ is a secant line to $K$ , and calculate the $x$ -coordinates and $y$ -coordinates of the intersection points $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$ of $h$ and $K$ .
$x_{1} =$

$y_{1} =$

$x_{2} =$

$y_{2} =$

Enter square roots as sqrt. Enter, for example, sqrt(2) for

\sqrt{2}

Substituting $g$ into $K$ results in

$x^{2} + (\sqrt{7} x + 5 - 1)^{2} = 2 \Leftrightarrow x^{2} + (\sqrt{7} x + 4)^{2} = 2 \Leftrightarrow x^{2} + 7 x^{2} + 8 \sqrt{7} x + 16 = 2$

$\Leftrightarrow 8 x^{2} + 8 \sqrt{7} x + 14 = 0 \Leftrightarrow 4 x^{2} + 4 \sqrt{7} x + 7 = 0 \Leftrightarrow (2 x + \sqrt{7})^{2} = 0 .$
Thus, the quadratic equation has exactly one solution, and $g$ is a tangent line to $K$ . For the $x$ -coordinate of the osculation point we have:

$(2 x + \sqrt{7})^{2} = 0 \Leftrightarrow 2 x + \sqrt{7} = 0 \Leftrightarrow x = - \frac{\sqrt{7}}{2} .$
Substituting $x$ into the given equation of a line results in:

$y = \sqrt{7} (- \frac{\sqrt{7}}{2}) + 5 = - \frac{7}{2} + 5 = \frac{3}{2} .$
Substituting $h$ in $K$ results in

$x^{2} + (2 x + 2 - 1)^{2} = 2 \Leftrightarrow x^{2} + (2 x + 1)^{2} - 2 = 0 \Leftrightarrow x^{2} + 4 x^{2} + 4 x + 1 - 2 = 0$

$\Leftrightarrow 5 x^{2} + 4 x - 1 = 0 \Leftrightarrow x_{1,2} = \frac{- 4 \pm \sqrt{16 + 20}}{10} = \frac{- 4 \pm 6}{10} \Leftrightarrow x_{1} = \frac{1}{5} \land x_{2} = - 1 .$
Since the quadratic equation has two solutions, $h$ is a secant line to $K$ . The corresponding $y$ -coordinates result again from substituting the $x$ -coordinates into the given equation of a line:

$y_{1} = 2 (\frac{1}{5}) + 2 = \frac{2}{5} + 2 = \frac{12}{5}$
and

$y_{2} = 2 (- 1) + 2 = - 2 + 2 = 0 .$

Exercise 9.3.13

Let a circle

K

be given by the equation

K : (x - 3)^{2} + y^{2} = 2,

and the line

g

with the slope

2

and the

y

-intercept

b

by the equation

g : y = 2 x + b

in normal form. Find the interval in which

b

must lie such that

g

is a secant line to

K

b \in]

;

[

.
Enter square roots as sqrt. Enter, for example, sqrt(2) as

\sqrt{2}

Substituting the equation of the line (with the unknown

b

) into the equation of a circle results in

(x - 3)^{2} + (2 x + b)^{2} = 2 \Leftrightarrow x^{2} - 6 x + 9 + 4 x^{2} + 4 b x + b^{2} = 2 \Leftrightarrow 5 x^{2} + (4 b - 6) x + b^{2} + 7 = 0 .

For

g

to be a secant line to

K

, this quadratic equation (in the variable

x

) must have two solutions. This is the case if its discriminant is positive, i.e. if we have:

(4 b - 6)^{2} - 4 \cdot 5 (b^{2} + 7) > 0 \Leftrightarrow 16 b^{2} - 48 b + 36 - 20 b^{2} - 140 > 0 \Leftrightarrow - 4 b^{2} - 48 b - 104 > 0

\Leftrightarrow b^{2} + 12 b + 26 < 0 .

The solutions of the quadratic equation

b^{2} + 12 b + 26 = 0

(in the variable

b

) are

b_{1,2} = \frac{- 12 \pm \sqrt{144 - 104}}{2} = \frac{- 12 \pm \sqrt{40}}{2} = \frac{- 12 \pm 2 \sqrt{10}}{2} = - 6 \pm \sqrt{10} .

Thus, the required interval is

] - 6 - \sqrt{10}; - 6 + \sqrt{10} [

Of course, lines defined by equations that cannot be transformed into normal form (i.e. lines that are parallel to the

y

-axis) can intersect or osculate circles as well. The method described above cannot be applied directly to such lines . The example below illustrates the method which is applied in this case.

Example 9.3.14

Let a circle

K

be given by the equation

K : (x - 1)^{2} + (y - 1)^{2} = 1,

and line

g

by the equation

g : x = \frac{3}{2} .

The two objects are shown in the figure below.

Obviously,

g

is a secant line to

K

. In this case, the intersection point cannot be calculated by solving the equation of a line for

y

. Instead, the equation of a line

x = \frac{3}{2}

is simply substituted into the given equation of a circle. This results in two values of

y

, i.e. the

y

-coordinates of the intersection points:

(\frac{3}{2} - 1)^{2} + (y - 1)^{2} = 1 \Leftrightarrow \frac{1}{4} + y^{2} - 2 y + 1 = 1 \Leftrightarrow y^{2} - 2 y + \frac{1}{4} = 0

\Leftrightarrow y_{1,2} = \frac{2 \pm \sqrt{4 - 1}}{2} = 1 \pm \frac{\sqrt{3}}{2} \Leftrightarrow y_{1} = 1 + \frac{\sqrt{3}}{2} \land y_{2} = 1 - \frac{\sqrt{3}}{2} .

Obviously, both corresponding

x

-coordinates are equal to

\frac{3}{2}

since the intersection points lie on the line

g

. Thus, the two intersection points of the line

g

and the circle

K

are

(\frac{3}{2}; 1 + \frac{\sqrt{3}}{2})

and

(\frac{3}{2}; 1 - \frac{\sqrt{3}}{2})

The Info Box below lists the different cases for the relative positions of two circles together with general criteria that allow to decide which of the cases for two given circles applies.

Info 9.3.15

Let two (different) circles,

K_{1}

with centre

M_{1}

and radius

r_{1}

and

K_{2}

with centre

M_{2}

and radius

r_{2}

, be given (by equations of a circle in the plane with respect to a fixed coordinate system). Then the circles have exactly one of the following three relative positions with respect to each other:

The circles $K_{1}$ and $K_{2}$ do not have any points in common, i.e. they do not intersect. This is the case if and only if for the radii $r_{1}$ and $r_{2}$ and the distance $[\overline{M_{1} M_{2}}]$ of the two centres, we have the inequalities

$[\overline{M_{1} M_{2}}] > r_{1} + r_{2} or [\overline{M_{1} M_{2}}] < | r_{1} - r_{2} | .$
The circles $K_{1}$ and $K_{2}$ have one point in common, i.e. they osculate. This is the case if and only if for the radii $r_{1}$ and $r_{2}$ and the distance $[\overline{M_{1} M_{2}}]$ of the two centres, we have the inequalities

$[\overline{M_{1} M_{2}}] = r_{1} + r_{2} or [\overline{M_{1} M_{2}}] = | r_{1} - r_{2} | .$
The circles $K_{1}$ and $K_{2}$ have two points in common, i.e. they intersect. This is the case if and only if for the radii $r_{1}$ and $r_{2}$ and the distance $[\overline{M_{1} M_{2}}]$ of the two centres, we have the inequality

$| r_{1} - r_{2} | < [\overline{M_{1} M_{2}}] < r_{1} + r_{2} .$

The three cases are illustrated in the figure below.

$[\overline{M_{1} M_{2}}] > r_{1} + r_{2}$ :	$[\overline{M_{1} M_{2}}] < \| r_{1} - r_{2} \|$ :

$[\overline{M_{1} M_{2}}] = r_{1} + r_{2}$ :	$[\overline{M_{1} M_{2}}] = \| r_{1} - r_{2} \|$ :

| r_{1} - r_{2} | < [\overline{M_{1} M_{2}}] < r_{1} + r_{2}

Thus, the relative position of two circles can be determined from their centres and radii. The calculation of possible intersection points is more difficult and goes beyond the scope of this course.

Example 9.3.16

Let a circle

K

with centre

M_{K}

and radius

r_{K}

be given by the equation

K : x^{2} + y^{2} - 2 y = 1,

and a circle

L

with centre

M_{L}

and unknown radius

r > 0

by the equation

L : (x + 1)^{2} + (y - 1)^{2} = r^{2} .

Find the values of

r

for which the circles

K

and

L

intersect, osculate, or do not intersect.
From the given equation of a circle

L

the centre

M_{L}

can be read off directly:

M_{L} = (- 1; 1)

. However, the equation for

K

is not in normal form. By completing the square this equation can easily be transformed into normal form such that the centre

M_{K}

and the radius

r_{K}

can be read off:

x^{2} + y^{2} - 2 y = 1 \Leftrightarrow x^{2} + y^{2} - 2 y + 1 - 1 = 1 \Leftrightarrow x^{2} + (y - 1)^{2} = 2 .

Thus, we have

M_{K} = (0; 1)

and

r_{K} = \sqrt{2}

. Hence, the distance between the two centres is

[\overline{M_{K} M_{L}}] = \sqrt{(0 + 1)^{2} + (1 - 1)^{2}} = 1,

and the sum of the two radii is

r + r_{K} = r + \sqrt{2} .

Since

r > 0

, we have

r + \sqrt{2} > \sqrt{2} > 1 = [\overline{M_{K} M_{L}}] .

Hence

[\overline{M_{K} M_{L}}] \geq r + \sqrt{2}

cannot occur in this case. The geometric reason for this is that the centre of

L

lies within

K

, as can be seen from the figure below. According to the criteria listed in Info Box 9.3.15, the two circles osculate if

| r - \sqrt{2} | = 1 = [\overline{M_{K} M_{L}}] .

According to the section on absolute value inequalities, this equation is satisfied if

r - \sqrt{2} = 1 or r - \sqrt{2} = - 1 .

Hence, the two radii

r

for which the circle

L

osculates the circle

K

are:

r = \sqrt{2} + 1 and r = \sqrt{2} - 1 .

Now we have still to investigate the remaining possible values of

r

. For

0 < r < \sqrt{2} - 1

, we have

| r - \sqrt{2} | = - (r - \sqrt{2}) = \sqrt{2} - r > 1 = [\overline{M_{K} M_{L}}],

i.e.

K

and

L

do not have any points in common. For

\sqrt{2} - 1 < r < \sqrt{2} + 1

, we have to distinguish the cases

\sqrt{2} - 1 < r \leq \sqrt{2}

and

\sqrt{2} < r < \sqrt{2} + 1

if we investigate the absolute value. In the case

\sqrt{2} - 1 < r \leq \sqrt{2}

, we have

| r - \sqrt{2} | = - (r - \sqrt{2}) = \sqrt{2} - r < 1 = [\overline{M_{K} M_{L}}],

and in the case

\sqrt{2} < r < \sqrt{2} + 1

, we have

| r - \sqrt{2} | = r - \sqrt{2} < 1 = [\overline{M_{K} M_{L}}] .

Thus, the circles

K

and

L

have two points in common for

\sqrt{2} - 1 < r < \sqrt{2} + 1

, i.e. the circles intersect. Finally, we have in the case

r > \sqrt{2} + 1

| r - \sqrt{2} | = r - \sqrt{2} > 1 = [\overline{M_{K} M_{L}}] .

The circles

K

and

L

also do not have any points in common in this case. In summary, we have the following conditions for the relative position of the two circles

K

and

L

If $r \in {\sqrt{2} - 1; \sqrt{2} + 1}$ , the two circles $K$ and $L$ osculate at one point.
If $r \in$ $] \sqrt{2} - 1; \sqrt{2} + 1 [$ , the two circles $K$ and $L$ intersect at two points.
If $r \in$ $] 0; \sqrt{2} - 1 [$ $\cup$ $] \sqrt{2} + 1; \infty [$ , the two circles $K$ and $L$ do not have any points in common.

The figure below shows some of the cases for the circles

K

and

L

Exercise 9.3.17

Let two circles

K_{1}

and

K_{2}

be given by the equations

K_{1} : (x + 6)^{2} + (y + 4)^{2} = 64

K_{2} : x^{2} + 2 x + y^{2} - 16 y + 40 = 0 .

The two circles

K_{1}

and

K_{2}

	osculate at one point,
	intersect at two points,
	do not have any points in common.

The centre

M_{1}

and the radius

r_{1}

of the circle

K_{1}

can immediately be read off:

M_{1} = (- 6; - 4)

and

r_{1} = 8

. To transform the equation of the circle

K_{2}

into normal form, we have to complete the square:

x^{2} + 2 x + y^{2} - 16 y + 40 = 0 \Leftrightarrow x^{2} + 2 x + 1 + y - 2 \cdot 8 y + 64 + 40 - 64 - 1 = 0

\Leftrightarrow (x + 1)^{2} + (y - 8)^{2} = 25 .

Now we can also read off the centre

M_{2}

and the radius

r_{2}

K_{2}

M_{2} = (- 1; 8)

and

r_{2} = 5

. From these values, we calculate

[\overline{M_{1} M_{2}}] = \sqrt{(- 1 + 6)^{2} + (8 + 4)^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13

and

r_{1} + r_{2} = 13 .

In this case we have

[\overline{M_{1} M_{2}}] = 13 = r_{1} + r_{2}

, and according to the criteria listed in Info Box 9.3.15, the circles

K_{1}

and

K_{2}

osculate at one point.

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics

Chapter 9 Objects in the Two-Dimensional Coordinate System

9.3.4 Relative Positions of Circles

Info 9.3.9

Info 9.3.10

Example 9.3.11

Exercise 9.3.12

Exercise 9.3.13

Example 9.3.14

Info 9.3.15

Example 9.3.16

Exercise 9.3.17