Tangents to a Circle
Theory
A tangent touches a circle at exactly one point and is perpendicular to the radius there. Two tangents from an external point are equal, and the angle between a tangent and a chord equals the angle in the alternate segment. This Year 10A maths page uses these facts to find angles and lengths and to write proofs.
A tangent is a line that touches a circle at exactly one point, the point of contact. At any point on a circle there is exactly one tangent, and it is perpendicular to the radius drawn to that point.
From a point \(P\) outside a circle you can draw two tangents. The tangent lengths \(PA\) and \(PB\), from \(P\) to the points of contact, are equal.
A chord drawn from the point of contact splits the circle into two segments. The alternate segment is the one on the other side of the chord from the angle you are looking at.
A circle drawn inside a triangle or quadrilateral so that it touches every side is inscribed; each side is a tangent. Two circles touch if they meet at one point and share a tangent there.
It prepares you for senior maths courses, where you prove results as well as use them.
For tangents \(PA\) and \(PB\) from \(P\) to a circle with centre \(O\):
The last result comes from the angle sum of quadrilateral \(OAPB\), which has two right angles. For a secant from \(P\) cutting the circle at \(A\) and \(B\), and a tangent \(PT\):
How to use tangent properties
- Mark a right angle wherever a radius meets a tangent, and equal ticks on the two tangents from each external point.
- Look for the shape: quadrilateral \(OAPB\) (angles add to \(360^\circ\)), isosceles triangle \(PAB\), right-angled triangle \(OAP\), or a chord from the point of contact (alternate segment).
- Calculate with a reason at every step.
Problem types
- Chains of tangents: when several tangents meet a circle, each external point gives a pair of equal lengths. Split each side into its two tangent lengths and add or subtract.
- Inscribed circle: write each side as a sum of two unknown tangent lengths, add the three equations, then subtract (Example 2).
- Lengths: use Pythagoras' theorem in the right-angled triangle radius-tangent-\(OP\). If the distance from \(P\) to the circle is given, \(OP = r + \text{distance}\) (Example 4). If an angle is given, use trigonometry.
- Tangent and secant from one point: \(PT^2 = PA \times PB\), where \(PB\) is the whole secant. If the parts are given in terms of \(k\), form an equation and keep the positive solution.
- Algebra with angles: set two expressions equal using a theorem (e.g. alternate segment), or use the angle sum of \(OAPB\).
- Proofs: join the centre to the points of contact and to \(P\). Right angles plus equal radii and a common hypotenuse suggest RHS congruence. For parallel lines, use the alternate segment theorem to find a pair of equal alternate angles. For two circles that touch, the common tangent at the point of contact gives equal tangents from one point to both circles.
(i) \(\angle AOB\)
(ii) \(\angle ACB\)
(iii) \(\angle PAB\).
(i) A tangent is perpendicular to the radius, so \(\angle OAP = \angle OBP = 90^\circ\). Angle sum of quadrilateral \(OAPB\):
| \(\angle AOB\) | \(=\) | \(360^\circ - 90^\circ - 90^\circ - 54^\circ\) |
| \(\) | \(=\) | \(\mathbf{126^\circ}\) |
(ii) Angle at the circumference is half the angle at the centre: \(\angle ACB = \mathbf{63^\circ}\).
(iii) \(PA = PB\) (tangents from an external point), so \(\triangle PAB\) is isosceles:
| \(\angle PAB\) | \(=\) | \(\dfrac{180^\circ - 54^\circ}{2} = \mathbf{63^\circ}\) |
\(\angle PAB = \angle ACB\), exactly as the alternate segment theorem says.
Tangents from an external point are equal: let \(AF = AE = p\), \(BF = BD = q\) and \(CD = CE = r\).
| \(p + q\) | \(=\) | \(11\) |
| \(q + r\) | \(=\) | \(14\) |
| \(p + r\) | \(=\) | \(9\) |
Add all three: \(2(p + q + r) = 34\), so \(p + q + r = 17\). Subtract each equation from this:
| \(p\) | \(=\) | \(17 - 14 = 3\) |
| \(q\) | \(=\) | \(17 - 9 = 8\) |
| \(r\) | \(=\) | \(17 - 11 = 6\) |
\(AF = \mathbf{3}\) cm, \(BD = \mathbf{8}\) cm, \(CE = \mathbf{6}\) cm. Check: \(8 + 6 = 14 = BC\).
The angle between a tangent and a chord equals the angle in the alternate segment:
| \(\angle ABC\) | \(=\) | \(\angle TAC = \mathbf{64^\circ}\) |
| \(\angle ACB\) | \(=\) | \(\angle SAB = \mathbf{50^\circ}\) |
Angles on the straight line \(SAT\):
| \(\angle BAC\) | \(=\) | \(180^\circ - 64^\circ - 50^\circ = 66^\circ\) |
Angle at the centre is twice the angle at the circumference on arc \(BC\):
| \(\angle BOC\) | \(=\) | \(2 \times 66^\circ = \mathbf{132^\circ}\) |
\(Q\) lies on \(OP\), so \(OP = r + 10\). The tangent is perpendicular to the radius, so \(\angle OTP = 90^\circ\). Pythagoras' theorem in \(\triangle OTP\):
| \((r + 10)^2\) | \(=\) | \(r^2 + 20^2\) |
| \(r^2 + 20r + 100\) | \(=\) | \(r^2 + 400\) |
| \(20r\) | \(=\) | \(300\) |
| \(r\) | \(=\) | \(\mathbf{15}\) |
Check: \(15^2 + 20^2 = 625 = 25^2\), and \(OP = 15 + 10 = 25\).
Common pitfalls
Frequently asked questions
Why is a tangent perpendicular to the radius?
The radius to the point of contact is the shortest distance from the centre to the tangent line, because every other point on the tangent is outside the circle. The shortest distance from a point to a line is the perpendicular.
Are two tangents from the same point equal?
Yes. Tangents drawn from an external point to the points of contact are equal in length. Joining the centre to the two points of contact and to the external point gives two congruent right-angled triangles (RHS).
What is the alternate segment theorem?
The angle between a tangent and a chord drawn from the point of contact equals the angle at the circumference in the alternate segment, the segment on the other side of the chord.
How do you find the radius from a tangent?
Draw the radius to the point of contact, which makes a right angle with the tangent. Then use Pythagoras' theorem in the triangle formed by the radius, the tangent and the line from the external point to the centre.
What is the tangent-secant theorem?
If a tangent \(PT\) and a secant through \(A\) and \(B\) are drawn from the same external point \(P\), then \(PT^2 = PA \times PB\).