Number of tangents from a point

Take a circle with centre $O$ and radius $r$, and a point $A$ anywhere in the plane. Compare the distance $OA$ with $r$ and only three outcomes are possible.

• $A$ inside the circle ($OA < r$): no tangent exists. Every line drawn through $A$ is forced to cross the circle boundary twice (a secant or a chord); none can touch it at a single point only.
• $A$ on the circle ($OA = r$): exactly one tangent exists — the unique line through $A$ perpendicular to the radius $OA$.
• $A$ outside the circle ($OA > r$): exactly two tangents exist, touching the circle at two distinct points, conventionally labelled $P$ and $Q$.

The interesting mathematics lives in the third case. If $AP$ and $AQ$ are the two tangents from an external point $A$, touching the circle at $P$ and $Q$, then the theorem you must prove states

where: $A$ is the external point, $P$ and $Q$ are the points of contact of the two tangents on the circle, and $O$ is the centre. The figure below shows the standard kite-shaped diagram $OPAQ$ that every proof and every worked problem in this topic is built around — line $OA$ splits the figure into two mirror-image right triangles.

Two facts feed directly into the proof. First, the radius-tangent theorem already known from earlier in the chapter: a tangent is always perpendicular to the radius drawn to the point of contact, so $\angle OPA = \angle OQA = 90^\circ$. Second, both $OP$ and $OQ$ are radii of the same circle, so $OP = OQ = r$. Those two facts, plus the shared side $OA$, are exactly the three pieces the RHS congruence test needs.

Problem. A point $A$ is at a distance of $13\text{ cm}$ from the centre $O$ of a circle of radius $5\text{ cm}$. Two tangents $AP$ and $AQ$ are drawn from $A$ to the circle. Find the length of each tangent, and verify that $AP = AQ$.

Step 1. Since $AP$ is a tangent at $P$, the radius-tangent theorem gives $OP \perp AP$, so triangle $OPA$ is right-angled at $P$ with hypotenuse $OA$. Apply Pythagoras' theorem:

Step 2. By the same argument at $Q$, triangle $OQA$ is right-angled at $Q$ with $OQ = OP = 5\text{ cm}$, so

Step 3. Since both computations used the identical values $OA = 13\text{ cm}$ and radius $5\text{ cm}$, the two tangent lengths come out equal, confirming

exactly as the theorem in the derivation above guarantees, without needing to measure or construct anything.

Final answer: each tangent has length $12\text{ cm}$, and $AP = AQ$ as the theorem predicts.

Competitive stretch

Problem. From the same external point $A$, if the two tangents $AP$ and $AQ$ meet at an angle $\angle PAQ = 70^\circ$, find $\angle POQ$, the angle between the two radii at the points of contact.

Step 1. In quadrilateral $OPAQ$, the two radii are each perpendicular to their respective tangents, so $\angle OPA = \angle OQA = 90^\circ$, and the angle sum of any quadrilateral is $360^\circ$:

Step 2. Substitute the known angles and solve:

The angle between the radii comes out to $\angle POQ = 110^\circ$, and notice it is the supplement of $\angle PAQ$ once the two right angles are removed from the quadrilateral.

• Trying to draw a tangent from a point that lies inside the circle — no tangent exists there; only chords and secants are possible.
• Forgetting that the right angle in the RHS congruence comes from the radius-tangent theorem, not from anything about the point $A$ itself — always mark $\angle OPA = \angle OQA = 90^\circ$ first.
• Assuming $AP = AQ$ applies to any two lines from $A$ touching the circle — the equal-length result is specific to the two tangents from one external point, not to arbitrary chords.
• Computing the Pythagoras step the wrong way round: since $OA$ is the hypotenuse, always use $AP^2 = OA^2 - OP^2$, never $OP^2 - OA^2$, which would give a negative (impossible) value.

Ask the tutor: Want to work through more tangent-length or tangent-angle problems, or see the RHS congruence proof redrawn step by step with a labelled diagram?