Theorem of the Line Perpendicular to the Radius and Tangent to a Circle

A line tangent to a circle is perpendicular to the circle's radius, and vice versa.

Consider a circle with center O and radius OP; the tangent line at point P is perpendicular to the radius.

the radius is perpendicular to the tangent line

The converse is also true: if a line is perpendicular to the radius at point P on a circle, then it is tangent to the circle at point P.

Note. This is because every line has a unique perpendicular passing through a given point P, and the shortest distance between a point and a line is always along the perpendicular. Therefore, the tangent line at point P on the circle is unique.

Proof

The proof is divided into two parts: one for the theorem and the other for its converse.

A] A Tangent Line at a Point on the Circle is Perpendicular to the Radius

First, let's prove that a line tangent to a circle is perpendicular to the radius.

Consider a circle with center O and radius OP, and a line tangent at point P on the circle.

example of a circle and a tangent line

By definition, a tangent line has only one point in common with the circle.

Knowing that through a point on a line, there is a unique perpendicular, we deduce that the tangent line has one and only one perpendicular at point P.

the radius is perpendicular to the tangent line

According to the theorem of the distance from a line to the center of a circle, a line is tangent to the circle at point P when the distance between the line (r) and the center (O) of the circle is equal to the radius (OP).

$$ d(r,O) \cong \overline{OP} $$

Since the (minimum) distance between a line and a point (O) is always perpendicular to the line, forming a right angle (90°), we deduce that the radius is perpendicular to the tangent line.

$$ \overline{OP} \perp r $$

Furthermore, we deduce that there is only one tangent line at point P.

B] A Line Perpendicular to the Radius at a Point on the Circle is Also Tangent to the Circle

Consider a circle with center O and radius OP.

Assume a line passes through point P and is perpendicular to the radius OP.

the radius is perpendicular to the tangent line

 

Knowing that every line has a unique perpendicular passing through a given point, we deduce there are no other lines at point P.

The shortest distance between a line and any point (O) is always a perpendicular segment (90°) to the line.

Therefore, segment OP measures both the distance between line r and point O and the radius.

the distance between a point and the line

According to the distance theorem, when the distance from a line to the center of the circle equals the radius, the line passes through only one point on the circle, making it a tangent line.

Therefore, the line perpendicular to the radius OP at its endpoint P is also a tangent line to the circle at the same point P.

How to Mathematically Verify That a Tangent Line Is Perpendicular to the Radius

To mathematically verify that a tangent line to a circle is perpendicular to the radius at the point of tangency, I can use analytical geometry.

Let me explain with a practical example.

Consider a circle centered at \( O(0, 0) \) with a radius of \( r = 5 \). The equation of the circle is:

$$ x^2 + y^2 = 25 $$

Suppose the tangent line has the equation:

$$ y = 3x + c $$

or equivalently,

$$ 3x - y + c = 0 $$

To find the point of tangency, we use the condition that the distance from the center of the circle to the tangent line must be equal to the radius of the circle (r = 5):

$$ d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} = r $$

$$ \frac{|3 \cdot 0 + (-1) \cdot 0 + c|}{\sqrt{3^2 + (-1)^2}} = 5 $$

$$ \frac{|c|}{\sqrt{9 + 1}} = 5 $$

$$ \frac{|c|}{\sqrt{10}} = 5 $$

Simplifying, we get:

$$ \frac{|c|}{\sqrt{10}} = 5 \quad \Rightarrow \quad |c| = 5\sqrt{10} $$

Therefore, \( c = 5\sqrt{10} \) or \( c = -5\sqrt{10} \).

For example, let's choose \( c = 5\sqrt{10} \), so the equation of the tangent line becomes:

$$ y = 3x + 5\sqrt{10} $$

To confirm, we can plot the circle and the tangent line:

example

The point of tangency \( T(x_1, y_1) \) satisfies both equations:

$$ \begin{cases} y_1 = 3x_1 + 5\sqrt{10} \\ \\ x_1^2 + y_1^2 = 25 \end{cases} $$

Substituting \( y_1 \):

$$ \begin{cases} y_1 = 3x_1 + 5\sqrt{10} \\ \\ x_1^2 + (3x_1 + 5\sqrt{10})^2 = 25 \end{cases} $$

$$ \begin{cases} y_1 = 3x_1 + 5\sqrt{10} \\ \\ x_1^2 + 9x_1^2 + 30\sqrt{10}x_1 + 250 = 25 \end{cases} $$

$$ \begin{cases} y_1 = 3x_1 + 5\sqrt{10} \\ \\ 10x_1^2 + 30\sqrt{10}x_1 + 225 = 0 \end{cases} $$

Solving this equation, we find \( x_1 \) and then \( y_1 \).

Solving the quadratic equation, we get \( x_1 = -1.5\sqrt{10} \approx -4.743 \)

$$ \begin{cases} y_1 = 3x_1 + 5\sqrt{10} \\ \\ x_1 = -1.5\sqrt{10} \approx -4.743 \end{cases} $$

Note: These are the steps to find the value of \( x_1 \). $$ x_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where \( a = 10 \), \( b = 30\sqrt{10} \), \( c = 225 \). $$ x_1 = \frac{-30\sqrt{10} \pm \sqrt{(30\sqrt{10})^2 - 4 \cdot 10 \cdot 225}}{2 \cdot 10} $$ $$ x_1 = \frac{-30\sqrt{10} \pm \sqrt{9000 - 9000}}{20} $$ $$ x_1 = \frac{-30\sqrt{10}}{20} $$ $$ x_1 = -1.5\sqrt{10} \approx -4.743 $$

Substituting \( x_1 = -1.5\sqrt{10} \) (approximately -4.743) into the first equation, we get \( y_1 = 0.5\sqrt{10} \) (approximately 1.58)

$$ \begin{cases} y_1 = 3 \cdot (-1.5\sqrt{10} ) + 5\sqrt{10} \\ \\ x_1 = -1.5\sqrt{10} \end{cases} $$

$$ \begin{cases} y_1 = -4.5\sqrt{10} + 5\sqrt{10} \\ \\ x_1 = -1.5\sqrt{10} \end{cases} $$

$$ \begin{cases} y_1 = 0.5\sqrt{10} \approx 1.58 \\ \\ x_1 = -1.5\sqrt{10} \approx -4.743 \end{cases} $$

Therefore, the point of tangency P is at coordinates (x1, y1) = (-1.5√10, 0.5√10) or approximately (-4.743, 1.58).

coordinates of the point of tangency

Now, we can find the equation of the radius passing through the center of the circle \( O(0, 0) \) and the point of tangency P.

The equation of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

$$ y - y_1 = m (x - x_1) $$

where \( m \) is the slope of the line, calculated as:

$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$

In this case, the points are \( O(0, 0) \) and \( T(-4.743, 1.58) \).

So, the slope \( m \) is:

$$ m = \frac{0.5\sqrt{10} - 0}{-1.5\sqrt{10} - 0} = \frac{0.5\sqrt{10}}{-1.5\sqrt{10}} = -\frac{1}{3} \approx -0.333 $$

Now, we can write the equation of the line passing through the radius and the point of tangency.

$$ y - 0 = - \frac{1}{3} (x - 0) $$

So the equation of the radius line is:

$$ y = - \frac{1}{3} x $$

For completeness, let's add the line \( y = -\frac{1}{3}x \) (red) that contains the radius OP in the graph.

the radius line

At this point, we need to verify that the radius is indeed perpendicular to the tangent.

Knowing that two lines are perpendicular if the product of their slopes is \( -1 \), we need to check:

$$ m_{\text{radius}} \cdot m_{\text{tangent}} = -1 $$

The slope of the tangent is \( 3 \) while that of the radius is \( -\frac{1}{3} \).

$$ -\frac{1}{3} \cdot 3 = -1 $$

The result is -1, so we can confirm that the radius is indeed perpendicular to the tangent.

This mathematically proves that any tangent line is perpendicular to the radius at the point of tangency.

And so on.

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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