Tangents and the Radius

A line passes through a point $T$ on a circle with centre $O$. Slide the line around $T$ like a see-saw. For most positions it cuts the circle again, giving a chord. There is exactly one position where it does not. At that position, what has happened to the distance from $O$ to the line, and what does that tell you about the angle between the line and $OT$?

At every point of a circle there is exactly one tangent, and it is perpendicular to the radius drawn to the point of contact. Everything else in this lesson is that right angle used somewhere.

Take a point $T$ on a circle with centre $O$ and radius $r$, and swing a line about $T$. For a general position the line cuts the circle at a second point as well, making a chord. Only one position avoids that.

The reason is the distance from $O$ to the line. In Lesson 2 you proved that the perpendicular from the centre bisects a chord, which means the distance $d$ from $O$ to a chord satisfies $d < r$. A line through $T$ has $d \leq r$ always, since $T$ is on the line and $OT = r$. So:

$$d < r \iff \text{the line is a secant} \qquad d = r \iff \text{the line is a tangent}$$

The distance from a point to a line is measured along the perpendicular, and there is exactly one line through $T$ whose perpendicular distance from $O$ equals $OT$ itself: the one perpendicular to $OT$. One perpendicular direction, one tangent.

To prove: if $t$ is a tangent at $T$ to a circle with centre $O$, then $OT \perp t$.

Proof by contradiction. Suppose $OT$ is not perpendicular to $t$. Let $F$ be the foot of the perpendicular from $O$ to $t$, so $F \neq T$. In the right triangle $OFT$ the hypotenuse is $OT$, so

$$OF < OT = r$$

That places $F$ inside the circle. But a line through an interior point must cut the circle twice, so $t$ is a secant, contradicting the assumption that it is a tangent.

Therefore no such $F$ exists, which means the foot of the perpendicular is $T$ itself, and $OT \perp t$.

The converse is also true, and is proved directly: if a line through $T$ is perpendicular to $OT$, then every other point $X$ on it satisfies $OX > OT = r$ (the hypotenuse of a right triangle exceeds its legs), so every other point is outside the circle and the line touches at $T$ alone.

Let $P$ be outside the circle, with tangents touching at $A$ and $B$.

To prove: $PA = PB$.

Join $OA$, $OB$ and $OP$. By the previous result $\angle OAP = \angle OBP = 90°$. Now compare $\triangle OAP$ and $\triangle OBP$:

$OA = OB$ (radii), $OP$ is common, and both triangles have a right angle at the point of contact. That is the RHS congruence test, so $\triangle OAP \equiv \triangle OBP$ and therefore $PA = PB$.

Pythagoras gives the same result and a formula with it:

$$PA = PB = \sqrt{OP^2 - r^2}$$

Congruence delivers two more equalities for free: $\angle APO = \angle BPO$, so $OP$ bisects the angle between the tangents, and $\angle AOP = \angle BOP$. The figure $OAPB$ is a kite, and $OP$ is its axis of symmetry.

A tangent at $T$ and a chord $TA$ make an angle between them. The chord splits the circle into two segments, and the angle sits against one of them. The theorem says that angle equals any angle at the circumference in the other segment, the alternate one.

To prove: $\angle ATX = \angle APT$, where $TX$ is the tangent on one side and $P$ is on the arc on the other side of $TA$.

Proof. Draw the diameter $TD$ from the point of contact. Then:

$$\angle DTX = 90° \quad \text{(tangent} \perp \text{radius)}$$

$$\angle TAD = 90° \quad \text{(angle in a semicircle)}$$

In $\triangle TAD$ the angles sum to $180°$, so $\angle ATD = 90° - \angle ADT$. And since $\angle DTX = 90°$,

$$\angle ATX = 90° - \angle ATD = \angle ADT$$

Finally $\angle ADT$ and $\angle APT$ both stand on the same arc $AT$, so they are equal by the same-segment corollary from Lesson 3. Chaining the two gives $\angle ATX = \angle APT$.

Four problems on the tangent results. Work each one, then reveal the answer.

1. 1 A circle has radius $5$ and $P$ is $13$ from the centre. Find the tangent length from $P$.

   Show answer
   Pythagoras on the right triangle: $\sqrt{13^2 - 5^2} = \sqrt{144}$.$12$

2. 2 Tangents from $P$ touch at $A$ and $B$, and $\angle AOB = 118°$. Find $\angle APB$.

   Show answer
   The quadrilateral $OAPB$ has two right angles, so $360° - 90° - 90° - 118°$.$62°$

3. 3 A tangent-chord angle is $37°$. Find the angle in the alternate segment.

   Show answer
   The alternate segment theorem makes them equal.$37°$

4. 4 $OT = 10$ is a radius and the tangent at $T$ meets $OP$ at $P$ with $\angle TOP = 60°$. Find $OP$.

   Show answer
   $\angle OTP = 90°$, so $\cos 60° = \dfrac{10}{OP}$.$OP = 20$