Tangents
Master tangents to a circle for Year 11 Specialist Mathematics in Queensland (QCAA). A tangent touches a circle at a single point, and three key results — the tangent is perpendicular to the radius, the two tangents from a point are equal, and the alternate segment theorem — unlock the angle and length work.
You will learn to find unknown angles and lengths and to justify each step with the right theorem, building the circle-geometry proof skills at the heart of the Unit 2 topic "Circle and geometric proofs".
Theory
A tangent touches a circle at exactly one point. This page covers the three tangent results used in Year 11 Specialist Mathematics (QCAA, Queensland): a tangent is perpendicular to the radius at the point of contact, the two tangents from an external point are equal, and the alternate segment theorem — with full worked examples.
A tangent to a circle is a line that touches the circle at exactly one point, the point of contact, without crossing inside it. A line that cuts the circle at two points is a secant, not a tangent.
Tangent perpendicular to the radius. The radius drawn to the point of contact is perpendicular to the tangent there: the angle between them is \(90^\circ\). This turns any tangent-and-radius picture into a right-angled triangle, so Pythagoras gives unknown lengths.
Two tangents from an external point. From a point outside the circle you can draw exactly two tangents, and they are equal in length. Joining them to the point of contact makes an isosceles triangle, so its base angles are equal.
Alternate segment theorem. The angle between a tangent and a chord drawn from the point of contact equals the angle in the alternate segment — the inscribed angle standing on that chord on the other side. Each of these results can be proved and is listed in the QCAA Unit 2 topic "Circle and geometric proofs".
With a tangent \(TP\) touching at \(P\) and centre \(O\) (radius \(r\)), the triangle \(OPT\) is right-angled at \(P\), so the tangent length is
For two tangents from \(T\) touching at \(A\) and \(B\), the quadrilateral \(OATB\) has two right angles, so its remaining angles satisfy
The alternate segment theorem, with the tangent-chord angle \(\angle TAB\) and \(C\) on the major arc, gives
How to solve a tangent problem
- Mark the right angles. Draw the radius to each point of contact; the tangent meets it at \(90^\circ\).
- Name equal lengths. Two tangents from the same external point are equal, so tick \(TA = TB\) and look for an isosceles triangle.
- Choose the theorem. Use Pythagoras for a length, the angle sum of \(\triangle\) or quadrilateral \(OATB\) for an angle, or the alternate segment theorem for a tangent-chord angle.
- Chain the steps and state the reason beside each line.
The tangent meets the radius at \(90^\circ\), so \(\triangle OPT\) is right-angled at \(P\); apply Pythagoras:
| \(\angle OPT\) | \(=\) | \(90^\circ\) |
| \(TP^2\) | \(=\) | \(OT^2 - OP^2\) |
| \(=\) | \(13^2 - 5^2\) | |
| \(=\) | \(169 - 25\) | |
| \(=\) | \(144\) | |
| \(TP\) | \(=\) | \(12\) |
\(TP = 12\) cm.
Each tangent is perpendicular to its radius, and the four angles of quadrilateral \(OATB\) add to \(360^\circ\):
| \(\angle OAT\) | \(=\) | \(\angle OBT = 90^\circ\) |
| \(\angle AOB\) | \(=\) | \(360^\circ - \angle OAT - \angle OBT - \angle ATB\) |
| \(=\) | \(360^\circ - 90^\circ - 90^\circ - 40^\circ\) | |
| \(=\) | \(360^\circ - 220^\circ\) | |
| \(=\) | \(140^\circ\) |
\(\angle AOB = 140^\circ\).
The tangent-chord angle equals the angle in the alternate segment, then use the angle sum of \(\triangle ABC\):
| \(\angle ACB\) | \(=\) | \(\angle TAB \quad(\text{alternate segment})\) |
| \(=\) | \(64^\circ\) | |
| \(\angle BAC\) | \(=\) | \(180^\circ - \angle ABC - \angle ACB\) |
| \(=\) | \(180^\circ - 71^\circ - 64^\circ\) | |
| \(=\) | \(45^\circ\) |
\(\angle BAC = 45^\circ\).
The alternate segment gives the inscribed angle \(\angle ACB\); the angle at the centre is twice the angle at the circumference on the same arc:
| \(\angle ACB\) | \(=\) | \(\angle TAB = 35^\circ\) |
| \(\angle AOB\) | \(=\) | \(2 \times \angle ACB\) |
| \(=\) | \(2 \times 35^\circ\) | |
| \(=\) | \(70^\circ\) |
\(\angle AOB = 70^\circ\).
Common pitfalls
Frequently asked questions
Why is a tangent perpendicular to the radius?
The shortest distance from the centre to the tangent line is along the radius to the point of contact, and the shortest segment from a point to a line is perpendicular to it. So the radius meets the tangent at \(90^\circ\).
How do you find the length of a tangent from an external point?
The radius, the tangent and the line to the centre form a right-angled triangle. If the radius is \(r\) and the distance to the centre is \(OT\), the tangent length is \(TP = \sqrt{OT^2 - r^2}\) by Pythagoras.
Are the two tangents from a point always equal?
Yes. The two tangents drawn from any external point to a circle are equal in length. Triangles \(OAT\) and \(OBT\) are congruent (RHS), so \(TA = TB\).
What is the alternate segment theorem?
The angle between a tangent and a chord at the point of contact equals the inscribed angle in the alternate segment — the angle standing on that chord from a point on the other arc.
How is the angle between two tangents related to the central angle?
For tangents touching at \(A\) and \(B\), the quadrilateral \(OATB\) has two right angles, so \(\angle ATB + \angle AOB = 180^\circ\).