Theorem on the Distance of a Line from the Center of a Circle
A line is secant, tangent, or external to a circle if its distance from the center of the circle is less than, equal to, or greater than the radius, respectively. And vice versa.
In other words, three scenarios can occur:
If a line is closer to the center of the circle than the radius \( d < r \), it is a secant line to the circle.
If a line touches the circle at exactly one point, its distance from the center is equal to the radius \( d = r \), making it a tangent line to the circle.
Finally, if the line is farther from the center than the radius \( d > r \), it does not intersect the circle at all, thus it is an external line to the circle.
The inverse theorem also holds, meaning a secant, tangent, or external line has a distance from the center that is respectively less than, equal to, or greater than the radius.
How to Determine Intersection Points Algebraically
In analytic geometry, to find out whether a line is secant, tangent, or external to a circle, you just need to solve a system comprising the equations of the circle and the line.
$$ \begin{cases} x^2 + y^2 + ax + bx + c = 0 \\ \\ a'x + b'y + c' = 0 \end{cases} $$
The solutions (x, y) of the system represent the coordinates of the intersection points between the line and the circle.
Based on the number of solutions, three scenarios can occur:
- 0 solutions: the line is external to the circle
- 1 solution: the line is tangent to the circle
- 2 solutions: the line is secant to the circle
This approach allows us to determine the number of intersection points algebraically, without needing to graph the line and the circle.
A Practical Example
Let's consider a circle and a line. We need to determine if the line is secant, tangent, or external to the circle.
We don't have the graph of the line and the circle; we only have their equations.
The equation of the circle is:
$$ x^2 + y^2 - 6x - 4y + 4 = 0 $$
The equation of the line is:
$$ x - y - 4 = 0 $$
We set up a system with both equations:
$$ \begin{cases} x^2 + y^2 - 6x - 4y + 4 = 0 \\ \\ x - y - 4 = 0 \end{cases} $$
Now, we need to check if the system has solutions and how many there are.
To solve the system, we use the substitution method: we solve for y in the second equation and substitute it into the first:
$$ \begin{cases} x^2 + (x - 4)^2 - 6x - 4(x - 4) + 4 = 0 \\ \\ y = x - 4 \end{cases} $$
$$ \begin{cases} x^2 + x^2 - 8x + 16 - 6x - 4x + 16 + 4 = 0 \\ \\ y = x - 4 \end{cases} $$
$$ \begin{cases} 2x^2 - 18x + 36 = 0 \\ \\ y = x - 4 \end{cases} $$
We can simplify the first equation by dividing both sides by two:
$$ \begin{cases} x^2 - 9x + 18 = 0 \\ \\ y = x - 4 \end{cases} $$
We study the discriminant Δ of the first equation:
$$ \Delta = b^2 - 4ac $$
$$ \Delta = (-9)^2 - 4 \cdot 1 \cdot 18 $$
$$ \Delta = 81 - 72 $$
$$ \Delta = 9 $$
The discriminant is positive, so the system has two solutions. We can already state that the line is secant to the circle because it intersects it at two distinct points.
Note: By studying the discriminant, we can immediately determine the number of solutions for the first equation in the system. Depending on the value of Δ:
- If Δ > 0, the equation has two distinct real solutions, meaning the line is secant to the circle.
- If Δ = 0, the equation has one double real solution, meaning the line is tangent to the circle.
- If Δ < 0, the equation has no real solutions, meaning the line is external to the circle.
What are the coordinates of the intersection points?
Since the discriminant is positive (excluding the case where the line is external), we can precisely calculate the coordinates (x, y) of the intersection points.
We solve the equation $ x^2 - 9x + 18 = 0 $:
$$ x = \frac{-b \pm \sqrt{\Delta}}{2a} $$
$$ x = \frac{-(-9) \pm \sqrt{9}}{2 \cdot 1} $$
$$ x = \frac{9 \pm 3}{2} = \begin{cases} x = \frac{9 - 3}{2} = 3 \\ \\ x = \frac{9 + 3}{2} = 6 \end{cases} $$
Therefore, the first equation of the system has two solutions: x = 3 and x = 6.
$$ \begin{cases} 2x^2 - 18x + 36 = 0 \\ \\ y = x - 4 \end{cases} $$
We substitute the values of x into the second equation of the system to find the corresponding values of y:
$$ y = x - 4 = \begin{cases} y = 3 - 4 = -1 \\ \\ y = 6 - 4 = 2 \end{cases} $$
The first solution of the system is the pair x = 3 and y = -1.
The second solution is the pair x = 6 and y = 2.
Therefore, the line intersects the circle at the points (3, -1) and (6, 2).
In conclusion, this method allows us to algebraically determine the relative position of the line with respect to the circle without needing to graph them.
The Proof
The proof is developed in three cases:
A] Distance of the Line Less Than the Radius
Consider a line \( r \) and a circle with center \( O \) and radius \( OA \).
The distance from the line to the center \( O \) of the circle is the segment \( OB \).
By initial hypothesis, the distance of the line from the center is less than the radius.
$$ \overline{OB} < \overline{OA} $$
On the line \( r \), draw a segment \( BC \) with length equal to the radius \( OA \).
Next, draw a segment \( OC \) that connects point \( C \) with the center of the circle.
This forms a right triangle \( OBC \).
Since it is a right triangle, we deduce that the hypotenuse \( OC \) is greater than the legs \( OB \) and \( BC \).
Particularly, we are interested in the fact that OC > BC
$$ \overline{OC} > \overline{BC} $$
Knowing that \( BC \cong OA \), we deduce that the segment \( OC \) is greater than the radius.
$$ \overline{OC} > \overline{OA} $$
Therefore, point \( C \) is outside the circle.
The segment \( BC \) has an internal point (B) and an external point (C) relative to the circle.
Thus, \( BC \) intersects the circle at least at one point \( D \).
Repeat the same procedure on the opposite side of the line.
On the line \( r \), draw a segment \( BE \) congruent with the radius \( OA \) of the circle.
Then draw the segment \( OE \) to form a right triangle \( OBE \).
Here too, the hypotenuse \( OE \) is greater than the legs \( OB \) and \( EB \).
$$ \overline{OE} > \overline{EB} $$
The segment \( EB \) has an internal point (B) and an external point (E) relative to the circle.
Therefore, \( EB \) intersects the circle in at least one point \( F \).
This proves that the line \( r \) intersects the circle at two distinct points \( D \) and \( F \).
Therefore, the line \( r \) is a secant line to the circle.
B] Distance of the Line Equal to the Radius
In this case, by initial hypothesis, the line \( r \) has a distance from the center \( O \) of the circle equal to the radius \( OA \).
Thus, the initial hypothesis is \( OB = OA \)
$$ \overline{OB} = \overline{OA} $$
On the line \( r \), draw a segment \( BC \) congruent with the radius \( OA \), meaning BC \cong OA
Then connect points \( O \) and \( C \).
The triangle \( OBC \) is a right triangle.
In a right triangle, the hypotenuse is always greater than the individual legs. Therefore, OC > BC.
$$ \overline{OC} > \overline{BC} $$
Since \( BC \cong OA \) is congruent with the radius, the segment \( OC \) is greater than the radius \( OA \).
$$ \overline{OC} > \overline {OA} $$
From this, we deduce that point \( C \) is outside the circle.
Thus, the segment \( BC \) has an internal point (B) and an external point (C) relative to the circle.
This means that it intersects the circle at least at one point \( B \).
Repeat the same procedure on the other side.
Construct a segment \( BD \) on the line \( r \) where \( BD \cong OA \) is congruent with the radius of the circle.
Then connect points \( O \) and \( D \).
The triangle \( OBD \) is a right triangle.
Thus, the hypotenuse \( OD \) is greater than the leg \( BD \).
$$ \overline{OD} > \overline{BD} $$
Consequently, the segment \( OD \) is greater than the radius \( OA \) since \( BD \cong OA \) by construction.
$$ \overline{OD} > \overline{OA} $$
From this, we deduce that point \( D \) is outside the circle.
The segment \( BD \) has an internal point (B) and an external point (D) relative to the circle. Therefore, it must intersect the circle at least at one point \( B \).
In this case, the intersection point is still \( B \).
This means that the line \( r \) intersects the circle at only one point.
Therefore, the line \( r \) is a tangent line to the circle.
C] Distance of the Line Greater Than the Radius
In this case, the initial hypothesis is that the line \( r \) has a distance \( OB \) from the center \( O \) of the circle greater than the radius \( OA \).
Thus, the initial hypothesis is:
$$ \overline{OB} > \overline{OA} $$
The distance between a point and a line is represented by the shortest perpendicular segment that connects the point to the line.
If this distance exceeds the radius, then no other segment connecting point \( O \) to any point \( C \) on the line can be shorter.
This is easily demonstrated by considering that any other point \( C \) on the line forms a right triangle \( OBC \) where the segment \( OC \) is the hypotenuse.
In a right triangle, the hypotenuse is always greater than each of the individual legs, so OC > OB in any situation.
$$ \overline{OC} > \overline{OB} $$
Knowing that \( OB > OA \) is greater than the radius of the circle by initial hypothesis, we deduce that \( OC \) is also greater than the radius \( OA \).
$$ \overline{OC} > \overline{OB} > \overline{OA} $$
Therefore, every point on the line \( r \) is outside the circle. There are no common points.
Consequently, the line \( r \) is an external line to the circle.
The Inverse Theorem
The distance of a line from the center of a circle is less than, equal to, or greater than the radius if, respectively, the line is secant, tangent, or external to the circle.
The Proof
To prove the inverse theorem, we proceed by contradiction.
Assume, for the sake of argument, that a line tangent to a circle does not have a distance from the center equal to the radius.
There are two scenarios to consider:
- The distance is less than the radius
In this case, there are two distinct points where the line intersects the circle. This would mean that the line is both secant and tangent, which is logically impossible.
- The distance is greater than the radius
In this situation, the line does not touch the circle at any point. However, this contradicts the definition of a tangent, which must touch the circle at exactly one point.
From these contradictions, we conclude that the initial assumption is incorrect. Therefore, the opposite is true.
In conclusion, we can assert with certainty that a tangent line has a distance from the center of the circle equal to the radius.
And so on.