Tangent length properties

A tangent touches a circle at exactly one point, and that single point of contact is where all its useful geometry lives. Draw the radius to that point and it meets the tangent at a perfect right angle — always, for every tangent to every circle. Stand outside the circle instead and draw two tangents to it, and something equally tidy happens: both tangent segments turn out to be exactly the same length, and the line from your external point to the centre splits the angle between them evenly. This section builds those two facts — tangent-radius perpendicularity and equal tangents from an external point — plus the companion rule for two circles that touch, and shows how a triangle or a quadrilateral hiding in the diagram is usually the fastest route to a numeric answer.

The core insight

Two tangents drawn from the same external point to the same circle are always equal in length, and this falls out directly from the tangent being perpendicular to the radius at the point of contact.

The core idea

Fact 1 — tangent is perpendicular to the radius. If $PT$ is tangent to a circle with centre $O$ at the point $T$ , the radius drawn to $T$ meets the tangent at a right angle:

OT \perp PT, \qquad \angle OTP = 90^\circ

where: $O$ is the centre, $T$ is the point of contact, and $P$ is any other point on the tangent line. This holds at every point of contact, with no exceptions, and it is the fact every other tangent result is built on.

Fact 2 — point of contact of two touching circles. If two circles touch each other, either externally (sitting outside one another) or internally (one inside the other), at a single point $T$ , then $T$ always lies on the straight line joining the two centres. Both circles' radii to $T$ are perpendicular to the same tangent line at $T$ , so both radii — and hence both centres — lie along one straight line through $T$ .

Fact 3 — equal tangents from an external point. If $PA$ and $PB$ are the two tangents drawn from an external point $P$ to a circle, touching it at $A$ and $B$ , then:

PA = PB

where: $P$ is the external point and $A$ , $B$ are the two points of contact. Beyond the equal lengths, the segment $OP$ (external point to centre) bisects both the angle between the tangents, $\angle APB$ , and the angle $\angle AOB$ at the centre — one line does two jobs at once.

Figure

A circle with centre O and two tangents PA and PB from an external point P showing the equal radii OA and OB perpendicular to the tangents and the line OP bisecting angle APB

Derivation

Derivation AWhy PA equals PB

The goal is to show, using only the perpendicularity of Fact 1 and the RHS congruence condition already known from Class IX triangle work, that the two tangent segments drawn from any single external point to a circle must be exactly equal in length, with no measurement needed. Consider a circle with centre $O$ , and an external point $P$ from which tangents touch the circle at $A$ and $B$ .

Step 1. Draw the two radii $OA$ and $OB$ and the two tangent segments $PA$ , $PB$ , then draw $OP$ to complete triangles $OAP$ and $OBP$ . Since $OA$ and $OB$ are radii of the same circle, $OA = OB$ .

Step 2. Apply Fact 1 at each point of contact: the radius is perpendicular to the tangent at that point, so $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ . Both triangles $OAP$ and $OBP$ are therefore right triangles.

Step 3. Compare the two right triangles $OAP$ and $OBP$ . They share the hypotenuse $OP$ (common to both), have equal sides $OA = OB$ (radii), and equal right angles at $A$ and $B$ . By the RHS (right angle–hypotenuse–side) congruence condition, $\triangle OAP \cong \triangle OBP$ .

Step 4. Congruent triangles have equal corresponding sides, so the sides opposite the equal angles at $O$ give $PA = PB$ . The same congruence also matches $\angle OPA$ with $\angle OPB$ and $\angle AOP$ with $\angle BOP$ , which is exactly why $OP$ bisects both $\angle APB$ and $\angle AOB$ — a single congruence proof delivers all three results at once. Verified: RHS congruence of triangles OAP and OBP with OA = OB (radii), OP common, and right angles at A, B gives PA = PB, angle OPA = angle OPB, and angle AOP = angle BOP.

Worked example

From an external point $P$ , two tangents $PA$ and $PB$ are drawn to a circle of radius $5\ \text{cm}$ with centre $O$ , such that $OP = 13\ \text{cm}$ . Find the length of each tangent, and then find $\angle AOB$ if it is also given that $\angle APB = 70^\circ$ .

Step 1. In right triangle $OAP$ , $\angle OAP = 90^\circ$ by Fact 1, with $OA = 5\ \text{cm}$ (radius) and $OP = 13\ \text{cm}$ (hypotenuse). Apply Pythagoras' theorem:

OP^2 = OA^2 + AP^2 \implies 13^2 = 5^2 + AP^2 \implies 169 = 25 + AP^2 \implies AP^2 = 144 \implies AP = 12\ \text{cm}

Step 2. By Fact 3 (equal tangents), $PB = PA = \mathbf{12\ cm}$ — no separate calculation is needed for the second tangent.

Step 3. Now use $\angle APB = 70^\circ$ to find $\angle AOB$ . The four points $O, A, P, B$ form a quadrilateral with $\angle OAP = \angle OBP = 90^\circ$ (both tangent-radius angles). The angle sum of a quadrilateral is $360^\circ$ :

\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ \implies \angle AOB + 90^\circ + 70^\circ + 90^\circ = 360^\circ

Step 4. Solve: $\angle AOB = 360^\circ - 250^\circ = \mathbf{110^\circ}$ . So the tangents are each $12\ \text{cm}$ long, and the angle they subtend at the centre is $110^\circ$ .

Common traps

Forgetting that the tangent-radius right angle sits at the point of contact, not at the external point or the centre — mislabeling which vertex holds $90^\circ$ derails the whole solution.
Applying $PA = PB$ to tangents drawn to different circles, or from different external points — the equality only holds for two tangents from one point to one circle.
Asserting that $OP$ bisects $\angle APB$ without justification — it is true, but it follows from the RHS congruence in the derivation above, not by inspection.
Missing that quadrilateral $OAPB$ has two built-in right angles — using its $360^\circ$ angle sum is often the fastest route to an unknown angle, as in the worked example.

Ask the tutor: Want a hand setting up the right triangle or the quadrilateral in a tangent-length problem — ask the AI tutor to sketch the diagram with you.

Why must the tangent-radius angle be exactly 90 degrees and not just close to it?

Because the tangent touches the circle at only one point, any other line through that point would have to cross into the circle's interior on one side, making it a secant rather than a tangent. The radius is the shortest possible distance from the centre to the tangent line, and the shortest distance from a point to a line is always along the perpendicular. So the 90 degree angle is not approximate — it is forced by the very definition of "touches at exactly one point."

Does the equal-tangents property work if the two tangents come from points outside the circle at different distances?

No. The equality $PA = PB$ is specifically about two tangents drawn from the same external point $P$ . If you instead take two different external points, each will have its own pair of equal tangents, but a tangent from one point is not generally equal to a tangent from the other point unless the two points happen to be equidistant from the centre. Always check that both tangent segments start at the identical external point before applying the rule.

How is the tangent-length property actually used later, beyond just finding lengths and angles?

It is the foundation for the ruler-and-compass construction of tangents from an external point, and it underlies power-of-a-point results used with secants and chords elsewhere in the Circles chapter. The RHS congruence argument in the derivation above is also a template you will reuse: whenever a diagram has a pair of right triangles sharing a hypotenuse and one equal side, congruence is usually the fastest way in.

Tangent length properties

The core idea

Derivation

Worked example

Common traps

Before this

Where this leads