Tangent perpendicular to radius

The first major theorem of this chapter is short, clean, and powerful: the tangent to a circle at any point is perpendicular to the radius drawn to the point of contact. This single fact, combined with the Pythagoras theorem, solves an enormous fraction of the board questions in this chapter.

Why is this theorem so useful? Because whenever you spot a tangent in a problem, you can drop the radius to the point of contact, mark a right angle there, and immediately invoke right-triangle reasoning. The tangent and the radius together form one leg and the perpendicular at every point of tangency.

Statement and proof

Theorem. Let a circle have centre $O$ and let $\ell$ be a tangent to it at the point $P$ . Then $OP \perp \ell$ .

Proof. We give the standard proof by contradiction. Suppose, for the sake of argument, that $OP$ is not perpendicular to $\ell$ . Drop the perpendicular from $O$ to $\ell$ and call its foot $Q$ . By assumption $Q \neq P$ , since $OQ \perp \ell$ but $OP$ is not perpendicular to $\ell$ .

Now consider $\triangle OQP$ . It is right-angled at $Q$ , so $OP$ is its hypotenuse and $OQ$ is one of its legs. The hypotenuse is the longest side of a right triangle, so

$OQ < OP.$

But $OP = r$ (the radius). So $OQ < r$ , which means $Q$ lies inside the circle (its distance from the centre is less than the radius). Now $Q$ is on the line $\ell$ , so $\ell$ enters the interior of the circle at $Q$ . But a line that enters the interior of a circle must intersect the circle again, meaning $\ell$ meets the circle in at least two points. This contradicts the definition of a tangent (one intersection point).

Hence our assumption is wrong; $OP \perp \ell$ . $\blacksquare$

Converse

The converse is equally useful: if a line through a point $P$ on a circle is perpendicular to the radius $OP$ , then the line is the tangent at $P$ .

Proof. Let $\ell$ pass through $P$ on the circle with $\ell \perp OP$ . Suppose $\ell$ meets the circle again at $Q$ . Then $OQ = OP = r$ , so $\triangle OPQ$ is isosceles. Drop the perpendicular from $O$ to $PQ$ at its midpoint $M$ ; then $\angle OMP = 90^\circ$ . But also $OP \perp \ell$ , so $\angle OPQ = 90^\circ$ . A triangle cannot have two right angles. So $Q$ does not exist, $\ell$ meets the circle only at $P$ , and $\ell$ is the tangent at $P$ .

This converse is sometimes the cleaner tool. If you have a perpendicular to a radius at the radius's endpoint, you can immediately call the line "tangent" and use Theorem 2 (equal tangents from an external point).

Standard configurations

Most board problems are built from one of these configurations:

Tangent + radius + external point. The radius, the tangent segment, and the line from the external point to the centre form a right triangle.
Two tangents from an external point. The two radii and the two tangent segments form a kite-shaped figure with two right angles.
Common tangent to two circles. External or internal common tangents form trapezoid-like figures with perpendicular radii.

In every case, the first reflex on seeing a tangent should be: draw the radius to the point of contact and mark the right angle.

Worked examples

Example 1. A tangent from a point $A$ at distance $13$ cm from the centre of a circle of radius $5$ cm has length:

$AT^2 + r^2 = OA^2 \Rightarrow AT^2 = 169 - 25 = 144 \Rightarrow AT = 12$ cm.

Example 2. A circle has centre $O$ and radius $r$ . A tangent at $P$ meets a line through $O$ at $A$ . If $\angle OAP = 30^\circ$ and $OA = 10$ cm, find $r$ .

In $\triangle OAP$ , right-angled at $P$ , $\sin 30^\circ = OP/OA = r/10 = 1/2 \Rightarrow r = 5$ cm.

Example 3. A point $P$ is $17$ cm from the centre of a circle, and the tangent from $P$ is $15$ cm. Find the radius.

$r^2 = 17^2 - 15^2 = 289 - 225 = 64 \Rightarrow r = 8$ cm.

Example 4. A tangent at point $A$ on a circle of centre $O$ meets the diameter extended at $B$ . If $\angle OAB = 90^\circ$ and $\angle AOB = 60^\circ$ , find the radius given $AB = 9$ cm.

In right $\triangle OAB$ , $\tan 60^\circ = AB/OA = \sqrt 3 \Rightarrow OA = 9/\sqrt 3 = 3\sqrt 3$ cm.

Example 5. In a circle of centre $O$ , a tangent at $P$ and a tangent at $Q$ meet at an external point $T$ . The line $OT$ has length $25$ cm and the radius is $7$ cm. Find the tangent length and the angle $\angle PTQ$ .

Tangent length: $\sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24$ cm.

In right $\triangle OPT$ : $\sin(\angle OTP) = 7/25 \Rightarrow \angle OTP \approx 16.26^\circ$ , so $\angle PTQ = 2 \times 16.26^\circ \approx 32.52^\circ$ . (Or: $\cos(\angle OTP) = 24/25$ , $\tan(\angle OTP) = 7/24$ .)

Try it yourself

State and prove the theorem of tangent-radius perpendicularity.
The tangent from a point $P$ to a circle of radius $7$ has length $24$ . Find $OP$ .
A tangent at $A$ to a circle of centre $O$ has $OA = 5$ , $AT = 12$ on the tangent. Find $OT$ .
From a point $P$ outside a circle of radius $r$ , two tangents are drawn. If the chord of contact has length $2c$ , the perpendicular from $O$ has length $\sqrt{r^2 - c^2}$ . Verify with a sketch.
Prove the converse: a line perpendicular to the radius at a point on a circle is tangent.
A tangent to a circle of radius $6$ from an external point makes an angle of $30^\circ$ with the line joining the point to the centre. Find that distance.
A circle of radius $r$ is inscribed in an equilateral triangle of side $a$ . Find $r$ in terms of $a$ .
From a point at distance $d$ from the centre, the angle subtended by the tangent length at $O$ is $\theta$ . Show $\cos\theta = r/d$ .
Prove: $\angle OAT + \angle OTA = 90^\circ$ for a tangent $AT$ with $A$ the point of contact.
A common tangent to two circles of radii $r$ and $R$ ( $R > r$ ), centres $d$ apart, has length $\sqrt{d^2 - (R-r)^2}$ (external). Why?
A line $\ell$ is tangent to a circle. From the centre, the foot of perpendicular to $\ell$ is the only point of $\ell$ inside or on the circle. Why?
Two circles touch externally at $P$ . Show that the tangent at $P$ to each circle is the same line.

Pitfalls / Insight

A frequent slip-up is to mark the right angle at the wrong vertex of the configuration , for instance, at the external point instead of at the point of contact. The right angle is always at the point of contact, where the tangent meets the radius.

A nice mental image: imagine the radius as a fishing line, the circle as a pond, and the tangent as a beam of light grazing the pond's edge. The beam is perpendicular to the line where it would dip into the water.