The tangent to a circle

A tangent to a circle is a line that has exactly one point in common with the circle. That point is the point of contact.

• In the drawing, the line $\displaystyle PQ$ is a tangent to the circle, and its point of contact is $\displaystyle A$.
• The line $\displaystyle CD$ cuts the circle at two points, so it is no tangent.
• Every other point of the tangent is outside the circle.

Theorem: A tangent to a circle is perpendicular to the radius at the point of contact.

• Every point of the tangent but $\displaystyle A$ is outside the circle – so it is farther from $\displaystyle O$ than the radius.
• So $\displaystyle OA$ is the shortest segment from $\displaystyle O$ to the tangent.
• The shortest segment from a point to a line is the perpendicular – so $\displaystyle OA$ is perpendicular to the tangent.

Theorem (the converse): A line perpendicular to a radius at its end is a tangent to the circle.

Theorem: If two tangents to a circle pass through a point outside it, the tangent segments – from the point to the points of contact – are equal.

Given: $\displaystyle PA$ and $\displaystyle PB$ are tangents to the circle with centre $\displaystyle O$, at $\displaystyle A$ and $\displaystyle B$.
To prove: $\displaystyle {PA}={PB}$.

The proof below draws $\displaystyle OP$ and shows that the triangles $\displaystyle OAP$ and $\displaystyle OBP$ are congruent:

• Theorem: The segment joining the centre of the circle to a point from which two tangents are drawn bisects the angle between the tangents. You write the proof yourself, in Exercise 2.
• Theorem: If two tangents to a circle are parallel, the segment joining their points of contact is a diameter. You write the proof yourself, in Exercise 4.

A circle’s radius is $\displaystyle 5$, and the point $\displaystyle P$ is $\displaystyle 13$ from its centre $\displaystyle O$.
$\displaystyle PA$ is a tangent to the circle at $\displaystyle A$. How long is $\displaystyle PA$?

The tangent is perpendicular to the radius, so the triangle $\displaystyle OAP$ is right-angled, and $\displaystyle OP$ is its hypotenuse. By Pythagoras’ theorem:

$\displaystyle PA$ is a tangent to the circle with centre $\displaystyle O$, at $\displaystyle A$.
The circle’s radius is $\displaystyle 6$, and $\displaystyle {OP}={10}$.

$\displaystyle PA$ and $\displaystyle PB$ are tangents to the circle with centre $\displaystyle O$, at $\displaystyle A$ and $\displaystyle B$.
Prove: $\displaystyle {\angle APO}={\angle BPO}$.

$\displaystyle PA$ and $\displaystyle PB$ are tangents to the circle with centre $\displaystyle O$, at $\displaystyle A$ and $\displaystyle B$.
$\displaystyle {\angle APB}={50^{\circ}}$.
Find $\displaystyle \angle AOB$.

The line $\displaystyle AP$ is a tangent to the circle with centre $\displaystyle O$ at $\displaystyle A$, and the line $\displaystyle BQ$ is a tangent to it at $\displaystyle B$.
$\displaystyle AP\parallel BQ$.
Prove: $\displaystyle {\angle AOB}={180^{\circ}}$, that is, $\displaystyle AB$ is a diameter.