Big picture
A tangent touches a circle at one point. The geometry of this contact explains a belt stretched around a wheel, a straight road beside a circular park and many proof problems. The chapter's two central results connect a tangent with a radius and compare tangents drawn from the same external point.
A sketch can suggest these relationships but cannot prove them. Always state which lines are tangent, identify their contact points and name the circle's centre. Then use perpendicular radii, congruent triangles and angle sums to justify every step rather than measuring a drawing.
Learning goals
Distinguish tangents, secants and lines outside a circle.
Prove that the radius at contact is perpendicular to a tangent.
Prove and apply equality of tangents from an external point.
Calculate tangent lengths and angles using right triangles.
Solve tangential-quadrilateral and incircle problems.
Present geometry proofs with explicit reasons and correctly matched vertices.
Core ideas
1. Lines relative to a circle
A secant intersects the circle at two distinct points. A tangent has exactly one common point, the point of contact. A line missing the circle has no common point. The radius is a segment from centre to circumference, not a tangent.
Let the radius be r and let d be the perpendicular distance from centre O to the line. If d < r, the line is a secant. If d = r, it is tangent. If d > r, it lies outside. The perpendicular is the shortest distance from a point to a line, making this classification possible.
Worked example. A circle has radius 6 cm. Lines at perpendicular distances 4 cm, 6 cm and 9 cm from its centre are respectively secant, tangent and outside. A line drawn almost touching in a rough diagram must still be classified by its mathematical distance or stated condition.
2. Radius–tangent perpendicularity: theorem and proof
Theorem: the tangent at any point of a circle is perpendicular to the radius through that point. Let line l touch a circle at T and O be the centre. We must prove OT ⟂ l.
Take any other point Q on l. Because l meets the circle only at T, Q lies outside the circle, so OQ > OT = r. Thus OT is the shortest segment from O to l. The shortest segment from a point to a line is perpendicular to that line. Therefore OT ⟂ l. This proof uses the unique contact and the shortest-distance property, not visual appearance.
Conversely, a line through T perpendicular to OT is tangent: every other point Q on the line satisfies OQ² = OT² + TQ² > r² and lies outside the circle. The perpendicular-radius condition therefore also helps recognise a tangent.
Worked example. A tangent at T meets a line through O at P. Since OT ⟂ PT, triangle OTP is right-angled at T. If angle OPT = 32°, angle POT = 180° − 90° − 32° = 58°. A common error is placing the right angle at P rather than at contact T.
3. Number of tangents from a point
From a point inside a circle, no tangent can be drawn: any full line through it must cross the boundary twice. From a point on the circle, there is exactly one tangent, perpendicular to the radius there. From an external point, two distinct tangents can be drawn.
Describe the standard diagram as follows: P is outside the circle with centre O, PA touches at A and PB touches at B. Join OA, OB and OP. The right angles are OAP and OBP. Keep the contact labels fixed throughout the proof; changing which point lies on which tangent causes mismatched triangles.
Example. If OP = 13 cm for a circle of radius 5 cm, P is external because 13 > 5. Two tangents exist. If OP were 5 cm, P would be on the circle and there would be only one tangent.
4. Equality of tangent lengths: theorem and proof
Theorem: the lengths of tangents drawn from an external point to a circle are equal. In the standard diagram, PA = PB.
In right triangles OAP and OBP, OA = OB because both are radii, OP is common, and the angles at A and B are right angles by the radius–tangent theorem. The triangles are congruent by RHS. Corresponding sides therefore give PA = PB.
The same congruence gives angle APO = OPB and angle AOP = POB. Thus OP bisects both the angle between the tangents and the corresponding central angle. These are consequences of the theorem's proof, not assumptions that may be inserted before proving it.
Worked example. If PA = 3x − 2 cm and PB = x + 8 cm, set 3x − 2 = x + 8. Then 2x = 10, so x = 5. Each tangent is 13 cm. Substituting back into both expressions checks the result. Equality applies to tangent segments from the same external point, not arbitrary tangents from different points.
5. Tangent lengths and Pythagoras
The radius to contact forms a right triangle with the centre-to-external-point segment. Hence PT² = OP² − OT², or PT = √(OP² − r²). OP is the hypotenuse because it is opposite the right angle at T.
Worked example. Radius = 5 cm and OP = 13 cm. PT² = 13² − 5² = 169 − 25 = 144. Therefore PT = 12 cm, taking the positive root because length is positive. Both tangents from P have this length.
Inverse example. A tangent is 15 cm and radius 8 cm. OP² = 15² + 8² = 225 + 64 = 289, so OP = 17 cm. This is distance to the centre. The shortest distance from P to the circle along OP is OP − r = 17 − 8 = 9 cm, a different requested quantity.
6. Angles between tangents
In quadrilateral OAPB, angles at A and B are each 90°. Its angle sum gives angle APB + angle AOB = 180°, where angle AOB is the central angle through the region facing P, usually the minor angle. Reflex angles require careful interpretation of the diagram.
Worked example. If the smaller angle between PA and PB is 64°, the corresponding minor central angle AOB = 180° − 64° = 116°. OP bisects the tangent angle, so angle APO = 32°; it also bisects the central angle, giving angle AOP = 58°. Triangle OAP checks: 90° + 32° + 58° = 180°.
If a tangent angle is 60°, triangle APB is isosceles because PA = PB, and its remaining angles are (180° − 60°)/2 = 60°. It is therefore equilateral. If PA = 14 cm, chord AB is also 14 cm. Do not assume the contact chord always equals a tangent; the 60° condition is essential.
7. Tangential quadrilaterals
A tangential quadrilateral has an incircle touching all four sides. Label vertices A, B, C, D in order and contact points E, F, G, H on AB, BC, CD, DA. Equality from each vertex gives AE = AH, BE = BF, CF = CG and DG = DH.
Add alternate sides: AB + CD = AE + BE + CG + DG. Replace equal tangent parts to obtain AH + BF + CF + DH = AD + BC. Therefore AB + CD = BC + AD. This necessary property applies when the quadrilateral actually has an incircle; do not infer existence from an unlabelled sketch.
Worked example. A quadrilateral circumscribes a circle and has AB = 9 cm, BC = 12 cm and CD = 8 cm. Then 9 + 8 = 12 + AD, so AD = 5 cm. Perimeter = 9 + 12 + 8 + 5 = 34 cm. A negative missing side would signal inconsistent data or a mistaken equation.
8. Triangle incircle applications
An incircle touches all three sides of a triangle. Tangent segments from each vertex are equal. If triangle side lengths are a = BC, b = CA and c = AB, and semiperimeter s = (a + b + c)/2, tangent lengths from A are s − a, from B s − b, and from C s − c.
Derivation: write the three tangent lengths as x, y, z. Then x + y = c, y + z = a, and z + x = b. Adding gives x + y + z = s. Hence x = s − a, with the other results similarly. This is algebra based on equal tangents, not a separate assumption.
Worked example. Let AB = 13 cm, BC = 14 cm and CA = 15 cm. Semiperimeter = (13 + 14 + 15)/2 = 21 cm. Tangents from A = 21 − 14 = 7 cm; from B = 21 − 15 = 6 cm; from C = 21 − 13 = 8 cm. Check each side: 7 + 6 = 13, 6 + 8 = 14, and 8 + 7 = 15.
A reliable approach to questions
Draw the centre, external point and all contact points.
Join the centre to contacts and mark justified right angles.
State equal radii, common sides and equal tangents only where applicable.
Choose RHS congruence, Pythagoras, angle sum or tangent-part equations.
Show the reason for every proof step and substitute numerical answers back.
Check that lengths are positive and the external point lies beyond the radius.
Connections that matter
Triangle congruence supplies the proof of equal tangent lengths.
Pythagoras converts the radius–tangent right angle into length calculations.
Linear equations solve tangent expressions and incircle contact lengths.
Common mistakes
Putting the right angle at the external point: it belongs at contact.
Equating tangents from different external points: the theorem requires a common point.
Treating a secant as a tangent: check the number of intersections.
Using the central angle as the tangent angle: the corresponding angles are supplementary.
Revision checklist
I can classify lines relative to a circle.
I can prove radius–tangent perpendicularity.
I can prove equal tangents by RHS congruence.
I can calculate tangent lengths and angles.
I can solve tangential-quadrilateral and triangle-incircle problems.
I can write proofs with explicit, correctly matched reasons.