Interaction of Line with Circle

When a line is placed near a circle, there are exactly three ways it can interact with the circle:

1. Intersecting at two distinct points: The line crosses the circle at two separate points.
2. Touching at exactly one point: The line just "kisses" the circle, meeting it at exactly one point. This is called the condition of tangency, and such a line is called a tangent.
3. Not intersecting at all: The line completely misses the circle, not touching or intersecting it at any point.

How can we precisely determine this behavior without actually drawing the figure?

The key lies in measuring the perpendicular distance \( d \) from the center of the circle to the given line, and comparing this distance to the radius \( r \) of the circle.

Mathematically, we have three simple conditions:

\[ \begin{align} d < r &\implies \text{Line intersects the circle at two distinct points.}\\[6pt] d = r &\implies \text{Line is tangent to the circle (touches at exactly one point).}\\[6pt] d > r &\implies \text{Line does not intersect or touch the circle at all.} \end{align} \]

Thus, by simply calculating the perpendicular distance from the circle’s center to the line and comparing it to the radius, we can determine the exact relationship between the line and the circle without the need for any drawing.

Line and Standard Circle

Consider a line with the general equation \( y = mx + c \) and a circle centered at the origin \( O(0,0) \) with equation \( x^2 + y^2 = a^2 \).

To analyze how this line interacts with the given circle, we measure the perpendicular distance from the circle's center \( O(0,0) \) to the line \( y = mx + c \). Recall that the perpendicular distance \( d \) from a point \((x_1,y_1)\) to a line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Here, we rewrite the given line in the form \( mx - y + c = 0 \). Then, the perpendicular distance \( d \) from the center \( O(0,0) \) to the line \( mx - y + c = 0 \) is:

\[ d = \frac{|m \cdot 0 - 1 \cdot 0 + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{1 + m^2}} \]

We now distinguish three cases clearly based on comparing \( d \) and the radius \( a \):

Case 1: Line intersects the circle at two distinct points

This occurs when the perpendicular distance is less than the radius \( a \):

\[ \frac{|c|}{\sqrt{1+m^2}} < a \]

Squaring both sides gives a simpler condition:

\[ \frac{c^2}{1+m^2} < a^2 \implies c^2 < a^2(1+m^2) \]

Hence, the line cuts the circle at two distinct points if:

\[ c^2 < a^2(1+m^2) \]

Case 2: Line is tangent to the circle

This scenario, called the condition for tangency, arises when the perpendicular distance from the center to the line equals exactly the radius \( a \):

\[ \frac{|c|}{\sqrt{1+m^2}} = a \]

Squaring again, we have the simplified form known as the condition for tangency:

\[ c^2 = a^2(1+m^2) \]

Case 3: Line does not intersect the circle at all

Finally, the line completely avoids the circle when the perpendicular distance is greater than the radius \( a \):

\[ \frac{|c|}{\sqrt{1+m^2}} > a \]

Squaring gives the condition:

\[ c^2 > a^2(1+m^2) \]

Condition of Tangency using Algebraic Methods

We can also derive the condition for tangency purely algebraically, without using any geometric interpretation of distance.

Let the equation of the circle be

\[ x^2 + y^2 = a^2, \]

and let the line be

\[ y = mx + c, \]

where \( m \) is the slope and \( c \) is the y-intercept.

To find the points of intersection, we substitute \( y = mx + c \) into the equation of the circle:

\[ x^2 + (mx + c)^2 = a^2. \]

Expanding and simplifying, we get:

\[ x^2 + m^2x^2 + 2mcx + c^2 = a^2 \implies (1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0. \tag{1} \]

This is a quadratic in \( x \). The line will touch the circle if and only if this quadratic has exactly one real solution, i.e., if it has a repeated root. This occurs precisely when the discriminant is zero.

The discriminant \( \Delta \) of (1) is:

\[ \Delta = (2mc)^2 - 4(1 + m^2)(c^2 - a^2). \]

Compute:

\[ \Delta = 4m^2c^2 - 4(1 + m^2)(c^2 - a^2). \]

Now set \( \Delta = 0 \) for tangency:

\[ 4m^2c^2 - 4(1 + m^2)(c^2 - a^2) = 0. \]

Simiplifying:

\[ 0 = c^2(1 + m^2) - m^2c^2 - a^2(1 + m^2) \implies c^2 = a^2(1 + m^2). \]

Hence, the condition for tangency is:

\[ \boxed{c^2 = a^2(1 + m^2)}. \]

Example

Determine whether the line \( y = 2x + 1 \) intersects, touches, or does not intersect the circle

\[ x^2 + y^2 - 2x + 4y + 1 = 0. \]

Solution:

To analyze the interaction, we first identify the center and radius of the circle. Comparing the given equation

\[ x^2 + y^2 - 2x + 4y + 1 = 0 \]

with the standard form \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we get:

\[ 2g = -2 \implies g = -1,\quad 2f = 4 \implies f = 2,\quad c = 1. \]

Therefore, the center of the circle is \( (1, -2) \), and the radius is:

\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - 1} = \sqrt{1 + 4 - 1} = \sqrt{4} = 2. \]

Now we find the perpendicular distance from the center \( (1, -2) \) to the line. The line is given as \( y = 2x + 1 \), which we rewrite as:

\[ 2x - y + 1 = 0. \]

Using the distance formula, the perpendicular distance from \( (x_0, y_0) = (1, -2) \) to the line \( Ax + By + C = 0 \) is:

\[ d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|2(1) + (-1)(-2) + 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|2 + 2 + 1|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}. \]

Thus the distance from the center to the line is \( \sqrt{5} \), and the radius is \( 2 \). Since:

\[ \sqrt{5} > 2 \]

we conclude that the perpendicular distance is greater than the radius.

\[ \implies \text{The line does not intersect the circle at any point.} \]

Example

Is the line \( 7x - 4y + 23 = 0 \) a tangent to the circle

\[ x^2 + y^2 - 5x - 4y - 1 = 0 \, ? \]

Solution:

We first identify the center and radius of the circle. Comparing with the standard form \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we get:

\[ 2g = -5 \implies g = -\frac{5}{2},\quad 2f = -4 \implies f = -2,\quad c = -1. \]

Hence, the center is:

\[ C = (-g, -f) = \left( \frac{5}{2},\ 2 \right) \]

and the radius is:

\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left( \frac{25}{4} \right) + 4 + 1} = \sqrt{ \frac{25 + 16 + 4}{4} } = \sqrt{\frac{65}{4}} = \frac{\sqrt{65}}{2}. \]

We now compute the perpendicular distance \( d \) from the center \( \left( \frac{5}{2},\ 2 \right) \) to the line \( 7x - 4y + 23 = 0 \). The formula gives:

\[ d = \frac{|7 \cdot \frac{5}{2} - 4 \cdot 2 + 23|}{\sqrt{7^2 + (-4)^2}} = \frac{|\frac{35}{2} - 8 + 23|}{\sqrt{49 + 16}} = \frac{|\frac{35 - 16 + 46}{2}|}{\sqrt{65}} = \frac{65}{2\sqrt{65}} = \frac{\sqrt{65}}{2}. \]

Thus, the perpendicular distance equals the radius:

\[ d = r = \frac{\sqrt{65}}{2} \]

\[ \implies \text{The line is tangent to the circle.} \]

Tangent to a Circle with Given Slope

Let a circle be given by the equation

\[ (x - \alpha)^2 + (y - \beta)^2 = a^2, \]

where the center is \( (\alpha, \beta) \) and the radius is \( a > 0 \). We seek to find the equation of a line tangent to this circle, whose slope is specified to be \( m \in \mathbb{R} \).

We claim that the equation of such a tangent is

\[ y - \beta = m(x - \alpha) \pm a\sqrt{1 + m^2}. \]

Proof:

Introduce new variables \( X = x - \alpha,\ Y = y - \beta \). This is a translation of the coordinate system, shifting the origin to the center of the circle. Under this transformation, the equation of the circle becomes

\[ X^2 + Y^2 = a^2. \tag{1} \]

We now wish to find the equation of a line tangent to this circle with slope \( m \) in the new coordinates. Any line with slope \( m \) in the \( (X, Y) \)-plane has the form

\[ Y = mX + c, \tag{2} \]

where \( c \in \mathbb{R} \) is the Y-intercept.

Using the condition for tangency:

\[ \frac{|c|}{\sqrt{1 + m^2}} = a \implies |c| = a\sqrt{1 + m^2} \implies c = \pm a\sqrt{1 + m^2}. \]

Substituting this value into equation (2), we obtain the equations of the two tangents in the translated frame:

\[ Y = mX \pm a\sqrt{1 + m^2}. \tag{3} \]

Now we revert back to the original coordinates by replacing \( X = x - \alpha,\ Y = y - \beta \). Substituting into (3) gives:

\[ y - \beta = m(x - \alpha) \pm a\sqrt{1 + m^2}. \]

Hence, the required tangents to the original circle with given slope \( m \) are:

\[ \boxed{y - \beta = m(x - \alpha) \pm a\sqrt{1 + m^2}}. \]

This completes the derivation.

Length of Chord Intercepted by a Line on a Circle

Let a line \( y = mx + c \) intersect the circle

\[ x^2 + y^2 = r^2 \]

at two distinct points \( A \) and \( B \). The segment \( AB \) is the chord intercepted by the line on the circle. We aim to compute the length of this chord.

Let the center of the circle be \( C(0,\ 0) \), and radius \( r \). Drop a perpendicular from the center to the chord \( AB \), meeting it at point \( N \). Since the perpendicular from the center to any chord bisects the chord, we have:

\[ AN = NB \quad \text{and} \quad AB = 2 \cdot AN. \]

Let \( d \) be the perpendicular distance from the center \( C \) to the line \( y = mx + c \). Then:

\[ d = \frac{|c|}{\sqrt{1 + m^2}}. \]

In the right-angled triangle \( \triangle CAN \), with hypotenuse \( r \) and leg \( d \), we apply the Pythagorean theorem:

\[ AN^2 = r^2 - d^2 \implies AN = \sqrt{r^2 - d^2}. \]

Hence, the length of the chord is:

\[ \boxed{AB = 2 \sqrt{r^2 - d^2}}. \]

This is a fundamental geometric result. It expresses the length of the intercepted chord in terms of the radius \( r \) of the circle and the perpendicular distance \( d \) from the center to the line. It is useful to remember:

\[ \boxed{\text{Chord length} = 2 \sqrt{r^2 - d^2}}. \]