Two tangents from an external point

The second great theorem of this chapter is just as short as the first and just as useful: the lengths of the two tangents drawn from an external point to a circle are equal. A glance at the symmetric "kite" shape this configuration makes is enough to convince you intuitively; the formal proof uses congruent right triangles.

This theorem has a rich set of corollaries that solve a wide variety of problems involving inscribed quadrilaterals, common tangents, and incircles. Together with Theorem 1, it forms the entire technical toolkit for Chapter 10.

Statement and proof

Theorem. Let $T$ be a point outside a circle of centre $O$ . Let the two tangent lines from $T$ touch the circle at $A$ and $B$ . Then $TA = TB$ .

Proof. Join $OA$ , $OB$ , and $OT$ . Then:

$OA = OB$ (both are radii).
$OT = OT$ (common side).
$\angle OAT = \angle OBT = 90^\circ$ (tangent-radius perpendicularity, Theorem 1).

So $\triangle OAT$ and $\triangle OBT$ are right triangles with equal hypotenuse and one equal leg. By the RHS (right-angle hypotenuse side) congruence rule, $\triangle OAT \cong \triangle OBT$ . Therefore the corresponding sides are equal: $TA = TB$ . $\blacksquare$

Corollaries

The same congruence gives more than equal tangent lengths:

$\angle ATO = \angle BTO$ : the line $OT$ bisects the angle $\angle ATB$ between the two tangents.
$\angle AOT = \angle BOT$ : the line $OT$ bisects the angle $\angle AOB$ between the two radii.
$OT \perp AB$ at the midpoint of $AB$ : the line of centres is the perpendicular bisector of the chord of contact.

These corollaries are independently useful in board problems, especially when finding lengths in the "kite" $OATB$ or proving concurrence.

Tangent length formula

Let $T$ be at distance $d$ from the centre, with $r$ the radius. The tangent length $\ell = TA = TB$ satisfies, by Pythagoras in $\triangle OAT$ :

$d^2 = r^2 + \ell^2 \Rightarrow \ell = \sqrt{d^2 - r^2}.$

Note that $\ell$ is real and positive only if $d > r$ , that is, $T$ must be strictly outside the circle. If $d = r$ , the "tangent" length is zero ( $T$ is on the circle). If $d < r$ , no tangent exists.

The chord of contact

The segment $AB$ joining the two points of contact is called the chord of contact from $T$ . Length of $AB$ : in the kite $OATB$ , the diagonal $AB$ is bisected perpendicularly by $OT$ . Let $M$ be the midpoint of $AB$ . Then $\triangle OAM$ is right-angled at $M$ with hypotenuse $r$ and leg $OM$ . Using similar triangles in $\triangle OAT \sim \triangle AMT$ (both right-angled, sharing $\angle T$ in $\triangle OAT$ and $\angle T$ in $\triangle AMT$ as part of the larger right triangle), the relation $OA \cdot AT = OT \cdot AM$ holds.

A clean computation: $AM = r\ell / d$ (where $\ell = \sqrt{d^2 - r^2}$ ), so $AB = 2r\ell/d = 2r\sqrt{d^2 - r^2}/d$ .

Quadrilateral circumscribing a circle

Theorem. If a quadrilateral $ABCD$ has all four of its sides tangent to a circle (inscribed circle), then $AB + CD = BC + DA$ .

Proof. Let the points of tangency on $AB, BC, CD, DA$ be $P, Q, R, S$ respectively. By the equal-tangents theorem from each vertex:

$AP = AS$ (from $A$ ).
$BP = BQ$ (from $B$ ).
$CQ = CR$ (from $C$ ).
$DR = DS$ (from $D$ ).

Now $AB + CD = AP + PB + CR + RD = AS + BQ + CQ + DS = (AS + DS) + (BQ + CQ) = AD + BC$ . $\blacksquare$

The converse is also true: if $AB + CD = BC + DA$ in a convex quadrilateral, it has an inscribed circle. (Not asked in proofs at this level.)

Worked examples

Example 1. From an external point $T$ , two tangents are drawn to a circle of radius $5$ . If the tangent length is $12$ , find $OT$ .

$OT^2 = r^2 + \ell^2 = 25 + 144 = 169 \Rightarrow OT = 13$ .

Example 2. In a circle of centre $O$ , two tangents from $T$ touch at $A$ and $B$ and meet at $T$ with $\angle ATB = 60^\circ$ . Find $\angle AOB$ .

In the kite $OATB$ , $\angle OAT = \angle OBT = 90^\circ$ , $\angle ATB = 60^\circ$ . Sum of angles in a quadrilateral: $360^\circ$ . So $\angle AOB = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ$ .

A useful relation: $\angle AOB + \angle ATB = 180^\circ$ whenever two tangents from an external point form a kite with the radii.

Example 3. A quadrilateral $ABCD$ is drawn to circumscribe a circle. If $AB = 6, BC = 7, CD = 4$ , find $DA$ .

By the theorem: $AB + CD = BC + DA \Rightarrow 6 + 4 = 7 + DA \Rightarrow DA = 3$ .

Example 4. Two concentric circles have radii $5$ and $3$ . A chord of the larger circle is tangent to the smaller. Find the chord length.

The perpendicular from the centre to the chord has length $3$ (the smaller radius), and the chord lies on the larger circle ( $r = 5$ ). Half-chord $= \sqrt{25 - 9} = 4$ . Full chord $= 8$ .

Example 5. In a right triangle of legs $a$ and $b$ and hypotenuse $c$ , the radius of the inscribed circle is $r = (a + b - c)/2$ . Derive it.

Set the circle tangent to all three sides; by equal tangents, the tangent lengths from each vertex are $s - a, s - b, s - c$ where $s = (a + b + c)/2$ . Sum of two tangents from $A$ and $B$ (the right-angle vertex is $C$ ) equals $c$ : $(s - a) + (s - b) = c \Rightarrow 2s - a - b = c \Rightarrow a + b - c = a + b - (2s - a - b) = 2(a + b - s) = 2 \cdot r$ (in this triangle the radius is the tangent length from the right-angle vertex, namely $s - c$ ). So $r = s - c = (a + b - c)/2$ .

Try it yourself

State the theorem of equal tangents and write its proof.
From an external point $P$ , two tangents to a circle of radius $4$ touch at $A$ and $B$ . If $PA = PB = 3$ , find $OP$ .
Two tangents from an external point make an angle of $60^\circ$ ; the radius is $r$ . Find the distance from the external point to the centre in terms of $r$ .
A quadrilateral $ABCD$ circumscribes a circle, with $AB = 9, BC = 12, CD = 11$ . Find $DA$ .
Prove that the line joining the external point to the centre of a circle bisects the chord of contact perpendicularly.
From a point $5$ cm above the centre of a circle of radius $3$ cm, two tangents are drawn. Find the tangent length.
Two tangents from $T$ to a circle of centre $O$ and radius $r$ make an angle $\theta$ . Show $r = OT \sin(\theta/2)$ .
In a circle of centre $O$ , a tangent at $A$ and a tangent at $B$ meet at $P$ . If $OP = 10$ , $OA = 6$ , find $\angle APB$ .
A circle is inscribed in a right triangle with legs $6$ and $8$ . Find the inradius.
Two circles touch externally; show that the line joining their centres passes through the point of contact.
A quadrilateral $ABCD$ has an inscribed circle; if $AB = BC = CD$ , prove $DA = AB$ .
Two tangents from an external point are perpendicular to each other. Show that the tangent length equals the radius.

Pitfalls / Insight

(1) The "equal tangents" theorem is about tangents from the same external point , not about any two tangents to a circle.

(2) In the kite $OATB$ , students sometimes forget that two angles are right angles. Always mark both. The relation $\angle AOB + \angle ATB = 180^\circ$ is a frequent free mark on board problems.

(3) For circumscribed quadrilaterals, the "opposite sides sum" relation is symmetric and easy , but only applies when all four sides are tangent to the circle. Verify before invoking.