Tangents From an External Point: Proof & Examples

Why two strings tied to the same peg always come out the same length

Stand outside a circular pond and throw two ropes so each just grazes the water's edge, and the two ropes, measured from your hand to the point they touch, come out exactly equal. That is the theorem of tangents from an external point, and it drops straight out of a single right angle at each contact point.

What Are Tangents From an External Point?

A tangent is a line that touches a circle at exactly one point without crossing it. From a point $P$ outside the circle, you can draw exactly two such tangents, touching the circle at two different points, say $A$ and $B$. The segments $PA$ and $PB$ are the tangent lengths from $P$.

The theorem is short: the two tangent lengths from an external point are equal, so $PA = PB$. This rests on the basic behaviour of a tangent to circles and on the fact that a tangent meets the radius at the point of contact at $90°$.

By the end you will be able to prove $PA = PB$, compute a tangent length from the centre-distance, and use the properties in exam problems.

Why Are the Two Tangents From an External Point Equal?

They are equal because the two right triangles formed, $OAP$ and $OBP$, are congruent. Each tangent meets its radius at $90°$, the two radii $OA$ and $OB$ are equal, and both triangles share the same hypotenuse $OP$. Congruent triangles have equal corresponding sides, so $PA = PB$.

The deeper reason is that the figure is symmetric about the line $OP$: reflect the whole picture in $OP$ and tangent $PA$ lands exactly on tangent $PB$. That mirror symmetry is what forces the two lengths to match.

How Do You Prove the Two Tangents Are Equal?

The standard proof uses the RHS congruence criterion (Right angle, Hypotenuse, Side).

Given: A circle with centre $O$; an external point $P$; tangents $PA$ and $PB$ touching at $A$ and $B$.

To prove: $PA = PB$.

Construction: Join $OA$, $OB$, and $OP$.

Proof, step by step:

$$\angle OAP = \angle OBP = 90° \quad (\text{radius} \perp \text{tangent at contact})$$

In triangles $OAP$ and $OBP$:

$$OA = OB \quad (\text{radii of the same circle})$$

$$OP = OP \quad (\text{common hypotenuse})$$

$$\angle OAP = \angle OBP = 90°$$

By the RHS congruence rule, $\triangle OAP \cong \triangle OBP$. Congruent triangles have equal corresponding parts, so:

$$PA = PB$$

The result is one of the standard circle theorems, and it depends on the wider idea of congruence in triangles.

What Are the Properties of Tangents From an External Point?

Once $\triangle OAP \cong \triangle OBP$, several more facts follow at once.

• Equal tangent lengths. $PA = PB$, the headline result.

• $OP$ bisects the angle between the tangents. $\angle APO = \angle BPO$, so the line to the centre splits $\angle APB$ in half.

• $OP$ bisects the angle between the radii. $\angle AOP = \angle BOP$, so $OP$ also splits $\angle AOB$ in half.

• The tangent length formula. In right triangle $OAP$, the tangent length is found by the Pythagoras theorem:

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

where $OP$ is the distance from the external point to the centre and $r$ is the radius. How do you find the length of a tangent from an external point? Use this formula: subtract the square of the radius from the square of the centre-distance, then take the square root.

• $OAPB$ is a kite. With $PA = PB$ and $OA = OB$, the quadrilateral $OAPB$ is a kite whose diagonal $OP$ is its axis of symmetry.

Examples of Tangents From an External Point

Example 1

From an external point $P$, tangent $PA = 12$ cm. Find the length of the other tangent $PB$.

By the equal-tangents theorem:

$$PB = PA = 12 \text{ cm}$$

The other tangent is $12$ cm.

Example 2

A point $P$ is $13$ cm from the centre of a circle of radius $5$ cm. A student writes the tangent length as $13 - 5 = 8$ cm. Find the mistake.

Subtracting the radius from the centre-distance treats the three lengths as if they lay on one straight line. They do not: the radius, the tangent, and the line to the centre form a right triangle, so the lengths obey Pythagoras, not simple subtraction.

Use the tangent length formula:

$$PA = \sqrt{OP^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$$

The tangent is $12$ cm, not $8$ cm. The right angle at the contact point is what the subtraction shortcut throws away.

Example 3

A tangent from $P$ has length $8$ cm and the circle's radius is $6$ cm. Find the distance $OP$ from $P$ to the centre.

Rearrange the formula $PA^2 = OP^2 - r^2$:

$$OP = \sqrt{PA^2 + r^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}$$

The external point is $10$ cm from the centre.

Example 4

Two tangents from $P$ meet the circle at $A$ and $B$, and $\angle APB = 60°$. Find $\angle APO$.

Since $OP$ bisects $\angle APB$:

$$\angle APO = \tfrac{1}{2}\times 60° = 30°$$

The line to the centre makes a $30°$ angle with each tangent.

Example 5

The two tangents from $P$ are each $9$ cm and $\angle APB = 90°$. Find the distance $AB$ between the two contact points.

Triangle $APB$ has $PA = PB = 9$ cm and a right angle at $P$, so by Pythagoras:

$$AB = \sqrt{PA^2 + PB^2} = \sqrt{9^2 + 9^2} = \sqrt{162} = 9\sqrt{2} \text{ cm}$$

The contact points are $9\sqrt{2}$ cm apart.

Example 6

From an external point, the tangent length is $24$ cm and $OP = 25$ cm. Find the radius, then state how a second circle sharing these two tangents would relate to the first.

From $PA^2 = OP^2 - r^2$:

$$r = \sqrt{OP^2 - PA^2} = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7 \text{ cm}$$

The radius is $7$ cm. Any second circle that both these tangent lines also touch would share tangents with the first, the situation studied under common tangents.

Where Is This Used?

The equal-tangents result matters wherever a symmetric grip on a round object is needed.

• Belts and pulleys. A belt leaving a pulley runs along a tangent; the equal-tangent symmetry keeps drive belts balanced around wheels.

• Cutting and clamping. A V-block or a pair of jaws gripping a cylindrical rod touches it along two tangent lines; equal tangents mean the rod self-centres between the jaws.

• Geometry of lenses and mirrors. Light grazing a circular boundary and the construction of tangent lines both rely on the right angle at the contact point.

The tangent-from-a-point construction is a classical result, formalised in the treatment of tangent lines to circles, and it is the geometric reason a chuck or collet centres a round workpiece so reliably.

Mistakes to Watch For

Mistake 1: Subtracting instead of using Pythagoras

Where it slips in: Finding a tangent length from the centre-distance and radius.

Don't do this: Compute $OP - r$ as the tangent length.

The correct way: The tangent, radius, and centre-line form a right triangle, so $PA = \sqrt{OP^2 - r^2}$. The rusher who subtracts straight across drops the right angle that makes the problem a triangle at all.

Mistake 2: Assuming any two tangents to a circle are equal

Where it slips in: Two tangents drawn from different external points.

Don't do this: Set two unrelated tangent lengths equal.

The correct way: The equal-tangents theorem applies only to two tangents from the same point. Tangents from different points have no reason to match. The memoriser who recalls "tangents are equal" without the "from one point" clause misapplies it constantly.

Mistake 3: Forgetting the right angle at the point of contact

Where it slips in: Setting up the triangle before marking the $90°$.

Don't do this: Treat $\angle OAP$ as unknown.

The correct way: The radius always meets the tangent at $90°$ at the contact point. Mark that right angle first; it is what powers both the congruence proof and the length formula.

Conclusion

• The two tangents from an external point to a circle are always equal in length: $PA = PB$.

• The proof pairs triangles $OAP$ and $OBP$ by RHS congruence, using the right angle where each radius meets its tangent.

• The line from the external point to the centre bisects both the angle between the tangents and the angle between the radii.

• The tangent length is $\sqrt{OP^2 - r^2}$, straight from Pythagoras in the right triangle.

• The most common mistake is subtracting the radius from the centre-distance instead of using Pythagoras.

To work tangent proofs through with a teacher, explore Bhanzu's geometry tutor, our high school math tutor sessions, or math classes online. To watch a trainer prove the equal-tangents theorem live, you can book a free demo class.

Read More

• Tangents and normals — how tangent and normal lines are defined and used.

• Inscribed angle theorem — another circle result built on isosceles radii.

• Secant of a circle — the line that cuts a circle at two points, the tangent's near cousin.

• Parts of a circle — centre, radius, chord, and the elements a tangent proof names.

• Equal and unequal chords — how chord lengths relate to their distance from the centre.