Chapter 10 — Circles

Recall that a circle is the set of all points in a plane at a fixed distance, the radius, from a fixed point, the centre. A chord is a segment joining two points on the circle, a diameter is a chord through the centre, and an arc is a part of the circle between two points. In this chapter the circle is studied together with straight lines drawn in its plane.

Take a circle with centre O and a line PQ in the same plane. Slide the line across the circle. Three situations occur, and only three.

• Non-intersecting line. The line and the circle have no common point. The distance of the line from the centre is greater than the radius.
• Secant. The line cuts the circle at two distinct points A and B. The distance of the line from the centre is less than the radius. The segment AB is a chord of the circle, so a secant is a chord extended indefinitely in both directions.
• Tangent. The line touches the circle at exactly one point. The distance of the line from the centre is equal to the radius. The common point is called the point of contact, and the line is said to touch the circle at that point.

The word tangent comes from the Latin tangere, to touch. A line cannot meet a circle in more than two points, because through three points on one line no circle can pass; hence the three cases exhaust all possibilities.

Everyday illustrations help. A cycle wheel rolling on a level road touches the road at one point at any instant, so the road is a tangent to the wheel. When a coin is placed touching a ruler on a table, the edge of the ruler is a tangent to the coin. The strings of a pulley system leave the pulley wheel along tangents. A stone whirled in a sling and then released flies along the tangent to its circular path, a fact that Newton later explained by the law of inertia.

An activity makes the definition concrete. Fix a wire across a circular disc and rotate it about a point on the disc's edge. Whenever the wire crosses the disc it meets the rim at two points; at exactly one position it meets the rim at that single fixed point only, and that is the tangent. Rotating further, it again cuts at two points. Thus at a given point of a circle there is one and only one tangent.

The examination asks the definitions in one-mark questions and the number of tangents from a point in a later topic; the deeper properties, perpendicularity and equal tangent lengths, follow.

The three-position picture also explains why a tangent is said to be the limiting case of a secant. Take a secant PQ cutting the circle at A and B. Keep A fixed and move the line so that B slides along the circle towards A. The chord AB shrinks. When B finally reaches A the two points of intersection coincide, the chord has vanished, and the line no longer crosses the circle; it merely touches it at A. In this position the secant has become the tangent at A. The same happens if the secant is moved parallel to itself away from the centre: the chord it cuts becomes shorter and shorter, and at the moment the distance from the centre equals the radius the chord has length zero and the line is a tangent. Move it further and the line leaves the circle altogether.

Two consequences follow immediately and are used without further comment.

• There is only one tangent at a point of a circle. Rotating a line about a point A of the circle, only one position touches the circle at A alone; every other position through A cuts the circle at a second point.
• The common point of a tangent and the circle is called the point of contact. It is a single point; a tangent has no chord.

The limiting idea also gives a first look at why the tangent is perpendicular to the radius, a result proved rigorously in the next topic. For any chord AB, the perpendicular from the centre O to AB bisects it. As B slides towards A, the foot of this perpendicular, the mid-point of AB, moves towards A too, so in the limit the perpendicular from O falls at A itself. That is, OA is perpendicular to the tangent at A.

Tangents have a physical meaning as directions of motion. A point moving on a circle has, at every instant, its instantaneous direction of motion along the tangent at its position. Sparks from a grinding wheel fly off along tangents; water drops from a spinning umbrella leave along tangents; a satellite in a circular orbit, if gravity were suddenly switched off, would move away along the tangent.

A related notion is the tangent to a curve other than a circle, which the Intermediate course defines through calculus in exactly this way, as the limiting position of a secant. The circle is the first and most important case.

Theorem 1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre O and a tangent XY to the circle at a point P.
To prove: OP ⊥ XY.
Construction: Take any point Q on XY other than P and join OQ.

Proof. Since XY is a tangent, it meets the circle only at P. So Q, which is a point of XY different from P, is not on the circle. Q cannot be inside the circle either, because then the line XY would pass through an interior point and would have to cut the circle at two points, making it a secant and not a tangent. Therefore Q lies outside the circle, and hence OQ is greater than the radius: OQ > OP. This holds for every point Q of XY other than P. So OP is the shortest of all the distances from O to points of the line XY. But the shortest distance from a point to a line is the perpendicular distance. Hence OP is perpendicular to XY.

The proof rests on two facts: a tangent has no interior point of the circle on it, and the perpendicular is the shortest segment from a point to a line. Learn to state both.

Remarks that follow.

• By this theorem, at any point on a circle there can be one and only one tangent, since only one line through P is perpendicular to OP.
• The line containing the radius through the point of contact is called the normal to the circle at that point.
• The converse is also true: a line drawn through the end point of a radius and perpendicular to it is a tangent to the circle. This gives the construction of a tangent at a given point.
• The theorem lets every problem with a tangent be converted into a right-triangle problem: join the centre to the point of contact, and a right angle appears.

Worked example. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find PQ. Since OP ⊥ PQ, triangle OPQ is right-angled at P, so PQ² = OQ² − OP² = 144 − 25 = 119, PQ = √119 cm.

Worked example. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius r satisfies r² = 25² − 24² = 625 − 576 = 49, r = 7 cm.

How many tangents to a circle can pass through a given point? The answer depends on where the point lies, and the board asks it directly.

• Point inside the circle. Every line through an interior point cuts the circle at two points, so it is a secant. No tangent can be drawn from a point inside the circle.
• Point on the circle. There is exactly one tangent, the line through the point perpendicular to the radius to that point.
• Point outside the circle. There are exactly two tangents. Rotating a line about an external point, it starts by missing the circle, touches it once, cuts it as a secant, touches it again on the other side, and misses it again.

For an external point P, the two tangents touch the circle at two points, say Q and R. The segment PQ, from the external point to the point of contact, is called the length of the tangent from P to the circle. The two tangents PQ and PR are symmetric about the line joining P to the centre O, and the next topic proves that they are equal in length.

The number of tangents can also be understood through distance. From a point P at distance d from the centre of a circle of radius r: if d < r no tangent, if d = r one tangent, if d > r two tangents, each of length √(d² − r²). As P moves farther away, the tangent length grows and the angle between the two tangents shrinks; as P approaches the circle, the two tangents close up and coincide when P reaches the circle.

Related counts that appear as one-mark questions: a circle can have infinitely many tangents, one at each of its points; a tangent intersects the circle at exactly one point; a line intersecting a circle at two points is a secant; the number of tangents parallel to a given line is two, one on either side of the centre; and the tangents at the two ends of a diameter are parallel, as shown later.

An interesting count: given two circles, the number of common tangents is 4 if the circles are separate, 3 if they touch externally, 2 if they intersect at two points, 1 if they touch internally, and 0 if one lies inside the other. These are handled in the last topic.

Worked example. From a point P, 13 cm from the centre of a circle of radius 5 cm, two tangents are drawn. Each has length √(169 − 25) = 12 cm. If the points of contact are Q and R, then OQPR is a quadrilateral with right angles at Q and R, and its area is twice the area of triangle OQP, that is, 2 × ½ × 5 × 12 = 60 cm².

Theorem 2. The lengths of tangents drawn from an external point to a circle are equal.

Given: A circle with centre O, an external point P, and two tangents PQ and PR from P touching the circle at Q and R.
To prove: PQ = PR.
Construction: Join OP, OQ and OR.

Proof. Since a tangent is perpendicular to the radius through the point of contact (Theorem 1), ∠OQP = 90° and ∠ORP = 90°. Now in the right triangles OQP and ORP:
OQ = OR (radii of the same circle),
OP = OP (common hypotenuse),
∠OQP = ∠ORP = 90°.
Therefore ΔOQP ≅ ΔORP by the RHS congruence rule. Hence PQ = PR (corresponding parts of congruent triangles).

The same congruence gives two further facts that are used as often as the theorem itself:

• ∠OPQ = ∠OPR, that is, the centre lies on the bisector of the angle between the two tangents; OP bisects ∠QPR.
• ∠POQ = ∠POR, that is, OP bisects the angle between the two radii ∠QOR.

Also, since PQ = PR, the triangle PQR is isosceles, so ∠PQR = ∠PRQ, and since OP bisects the vertical angle it is also the perpendicular bisector of the chord QR; the chord of contact QR is perpendicular to OP.

An alternative proof uses the Pythagoras theorem: PQ² = OP² − OQ² and PR² = OP² − OR², and OQ = OR, so PQ² = PR² and PQ = PR. The congruence proof is preferred in the examination because it also delivers the angle bisector results.

Worked example. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove ∠PTQ = 2∠OPQ. Let ∠PTQ = θ. Since TP = TQ, ∠TPQ = ∠TQP = (180° − θ)/2 = 90° − θ/2. As OP ⊥ TP, ∠OPQ = 90° − ∠TPQ = θ/2. Hence ∠PTQ = 2∠OPQ.

Worked example. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at 80°, find ∠POA. OP bisects ∠APB, so ∠APO = 40°; in right triangle OAP, ∠POA = 90° − 40° = 50°.

A result asked in almost every board paper is the relation between the angle between two tangents from an external point and the angle subtended at the centre by the segment joining their points of contact.

Theorem. The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Given: Tangents PQ and PR from P touching the circle with centre O at Q and R.
To prove: ∠QPR + ∠QOR = 180°.

Proof. By Theorem 1, ∠OQP = 90° and ∠ORP = 90°. In quadrilateral OQPR, the sum of the angles is 360°, so ∠QOR + ∠OQP + ∠QPR + ∠ORP = 360°, that is, ∠QOR + 90° + ∠QPR + 90° = 360°. Hence ∠QOR + ∠QPR = 180°.

So the quadrilateral formed by the centre, the two points of contact and the external point has two opposite right angles, and its other two angles are supplementary; it is therefore a cyclic quadrilateral, with OP as a diameter of its circumcircle.

Worked example. If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, find ∠PTQ. ∠PTQ = 180° − 110° = 70°.

Worked example. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠AOB = 100°, and since OP bisects it, ∠POA = 50°.

Worked example. Two tangents from P to a circle of radius r are perpendicular to each other. Find the length of each tangent and OP. ∠QPR = 90°, so ∠QOR = 90° and quadrilateral OQPR has four right angles with OQ = OR = r; it is a square. Hence each tangent equals r and OP = r√2.

Worked example. The angle between two tangents from P is 60°. Then ∠QOR = 120°, ∠OPQ = 30°, and in right triangle OQP, OQ/OP = sin 30° = 1/2, so OP = 2r and the tangent length is PQ = OP cos 30° = r√3. The chord of contact QR subtends 120° at the centre, so QR = 2r sin 60° = r√3 as well; triangle PQR is equilateral.

Remember the chain: two right angles at the points of contact, angle sum 360° in the quadrilateral, hence supplementary. This four-line proof earns full marks when the figure and the two right angles are stated.

Result. The tangents drawn at the ends of a diameter of a circle are parallel.

Proof. Let AB be a diameter of a circle with centre O, and let PQ and RS be the tangents at A and B. By Theorem 1, OA ⊥ PQ and OB ⊥ RS, so ∠OAQ = 90° and ∠OBR = 90°. Now A, O, B are collinear, so AB is a transversal of PQ and RS, and the alternate angles ∠QAB and ∠ABR are both 90°, hence equal. Therefore PQ ∥ RS.

The converse also holds: if two parallel tangents are drawn to a circle, the points of contact are the ends of a diameter, since the radii to them are both perpendicular to the same direction and hence lie along one line through O.

Result. The perpendicular at the point of contact to the tangent passes through the centre. Since the tangent at P is perpendicular to OP, and only one line through P can be perpendicular to the tangent, that line is the one containing OP, which passes through O. This is the standard proof that the line through the point of contact perpendicular to the tangent contains the centre, and it is used to locate the centre of a circle from two tangents.

Worked example. Prove that the line segment joining the points of contact of two parallel tangents is a diameter. Let the tangents at A and B be parallel. OA ⊥ tangent at A and OB ⊥ tangent at B; two lines perpendicular to parallel lines are themselves parallel, but OA and OB share the point O, so they lie on one line; hence A, O, B are collinear and AB is a diameter.

Worked example. A pair of parallel tangents is drawn to a circle of radius 4 cm. The distance between them is the diameter, 8 cm. A third tangent cuts the two parallel tangents at C and D; if it touches the circle at E and the parallel tangents touch at A and B, then CA = CE and DB = DE (equal tangents from C and from D), so CD = CA + DB. Also ∠COD = 90°, since OC and OD bisect the supplementary angles ∠ACD and ∠BDC between the parallel lines.

Worked example. Two concentric circles have radii 5 cm and 3 cm. A chord of the larger circle touches the smaller circle at P. The radius OP is perpendicular to the chord, so it bisects the chord; half the chord is √(25 − 9) = 4 cm and the chord is 8 cm long. This is a much-asked application of the perpendicularity theorem together with the chord-bisection property of Class 9.

The results of this topic are short but frequently appear as two-mark proofs and as steps in longer problems on circumscribed figures.

A quadrilateral all of whose sides touch a circle is said to circumscribe the circle, and the circle is inscribed in the quadrilateral. Equal tangent lengths give a simple relation between the sides.

Theorem. If a quadrilateral ABCD circumscribes a circle, then AB + CD = AD + BC.

Given: Quadrilateral ABCD whose sides AB, BC, CD, DA touch a circle at P, Q, R, S respectively.
To prove: AB + CD = AD + BC.

Proof. Tangents from an external point are equal. From A: AP = AS. From B: BP = BQ. From C: CR = CQ. From D: DR = DS. Adding all four equations,
AP + BP + CR + DR = AS + BQ + CQ + DS,
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ),
AB + CD = AD + BC.

The sums of the two pairs of opposite sides are equal; the perimeter is twice either sum.

Corollary: a parallelogram circumscribing a circle is a rhombus. In a parallelogram AB = CD and AD = BC. Combining with AB + CD = AD + BC gives 2AB = 2AD, so AB = AD, and all four sides are equal; the parallelogram is a rhombus. Similarly a rectangle circumscribing a circle is a square.

Worked example. A quadrilateral ABCD is drawn to circumscribe a circle with AB = 6 cm, BC = 7 cm, CD = 4 cm. Find AD. AD = AB + CD − BC = 6 + 4 − 7 = 3 cm.

Worked example. Prove that the angle subtended at the centre by a side of the circumscribed quadrilateral and the angle subtended by the opposite side are supplementary, that is, ∠AOB + ∠COD = 180°. Join O to P, Q, R, S and to the vertices. OA bisects ∠SOP, OB bisects ∠POQ, OC bisects ∠QOR, OD bisects ∠ROS (the centre lies on the bisector of the angle between the radii to the points of contact of the two tangents from each vertex). So ∠AOB = ½(∠SOP + ∠POQ) and ∠COD = ½(∠QOR + ∠ROS); adding, ∠AOB + ∠COD = ½ × 360° = 180°.

Worked example. A circle is inscribed in a quadrilateral whose sides are 5 cm, 7 cm, 8 cm and x cm in order. Then 5 + 8 = 7 + x, so x = 6 cm.

The method to keep in mind: mark the four points of contact, write the four pairs of equal tangents, and add. The same idea, applied to a triangle, gives the incircle results of the next topic.

A circle that touches all three sides of a triangle is its incircle, with centre the incentre, the point where the three angle bisectors meet, and radius the inradius r. The equal-tangent theorem gives the lengths from each vertex to the points of contact.

Let the incircle of triangle ABC touch BC at D, CA at E and AB at F. Let a = BC, b = CA, c = AB and s = (a + b + c)/2, the semi-perimeter. From A: AE = AF = x; from B: BD = BF = y; from C: CD = CE = z. Then c = x + y, a = y + z, b = z + x, and adding, x + y + z = s. Hence

AF = AE = s − a, BD = BF = s − b, CD = CE = s − c.

Worked example. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm. Find AB and AC.
Let AF = AE = x. Then AB = x + 8, AC = x + 6, BC = 14. Area of ABC = area of OBC + OCA + OAB = ½ × 4 × (14 + x + 6 + x + 8) = 2(28 + 2x) = 56 + 4x. By Heron's formula with s = 14 + x, area = √[(14 + x)(x)(8)(6)] = √[48x(14 + x)]. So (56 + 4x)² = 48x(14 + x), 16(14 + x)² = 48x(14 + x), (14 + x) = 3x, x = 7. Hence AB = 15 cm and AC = 13 cm.

Area and inradius. Joining the incentre to the vertices splits the triangle into three triangles with heights r on bases a, b, c, so Area = ½ r(a + b + c) = rs. Thus r = Area/s.

Right triangle. For a right triangle with legs a, b and hypotenuse c, the inradius is r = (a + b − c)/2. Reason: the tangent lengths from the right-angle vertex are both r (the radii to the points of contact together with the two sides form a square), so a − r and b − r are the other tangent lengths and their sum is the hypotenuse: (a − r) + (b − r) = c. For the 3-4-5 triangle, r = (3 + 4 − 5)/2 = 1.

Worked example. The incentre of a right triangle with legs 6 and 8 is at distance r = (6 + 8 − 10)/2 = 2 from each side, and the area check gives rs = 2 × 12 = 24 = ½ × 6 × 8.

Worked example. In a triangle with sides 7, 8, 9 cm, s = 12, so the tangent lengths from the vertices opposite sides 7, 8, 9 are 5, 4, 3 cm respectively.

The board sets three- and four-mark proofs that combine the two theorems with congruence, isosceles triangles and angle chasing. Here are the standard ones with their proofs, each of which you should be able to reproduce.

Problem 1. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord. Let the tangents at A and B of chord AB meet at P. Then PA = PB (tangents from P), so triangle PAB is isosceles and ∠PAB = ∠PBA.

Problem 2. Prove that the perpendicular at the point of contact to the tangent passes through the centre. Let the tangent at P be XY and suppose the perpendicular to XY at P does not pass through O; join OP. Then OP ⊥ XY by Theorem 1, and the constructed perpendicular is also ⊥ XY at P, so two distinct lines through P are perpendicular to XY, which is impossible. Hence the perpendicular passes through O.

Problem 3. A circle touches the sides BC, CA, AB of triangle ABC at D, E, F. If the circle's centre is O, prove ∠BOC = 90° + ½∠A. This is a Class 9 style angle chase: OB and OC bisect ∠B and ∠C, so ∠BOC = 180° − ½(∠B + ∠C) = 180° − ½(180° − ∠A) = 90° + ½∠A.

Problem 4. Prove that the tangents at the ends of a diameter are parallel: done above via the two right angles and the transversal AB.

Problem 5. In two concentric circles, prove that a chord of the larger circle which touches the smaller circle is bisected at the point of contact. The radius OP to the point of contact is perpendicular to the chord (Theorem 1), and a perpendicular from the centre bisects a chord (Class 9), so the chord is bisected at P.

Problem 6. XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects XY at A and X′Y′ at B. Prove ∠AOB = 90°. Join O to the points of contact P on XY and Q on X′Y′ and to C. Triangles OPA and OCA are congruent (RHS: OP = OC, OA common, right angles), so ∠POA = ∠COA. Similarly ∠QOB = ∠COB. Since P, O, Q are collinear (tangents at the ends of a diameter), ∠POC + ∠COQ = 180°, hence 2∠COA + 2∠COB = 180°, so ∠AOB = ∠COA + ∠COB = 90°.

Problem 7. Prove that the angle between two tangents from an external point is supplementary to the angle between the radii: the quadrilateral angle-sum argument of the earlier topic.

Problem 8. Prove that a parallelogram circumscribing a circle is a rhombus: AB + CD = AD + BC with AB = CD and AD = BC gives AB = AD.

In each proof, name the theorem used at the step where you use it: tangent ⊥ radius; equal tangents from an external point; RHS congruence; angle sum property.

Numerical questions on this chapter combine the right angle at the point of contact with the Pythagoras theorem, the equal tangents, and sometimes similar triangles. Here are the patterns with full working.

Pattern 1: tangent length. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm; find the radius. r² = 25² − 24² = 49, r = 7 cm.

Pattern 2: chord of contact. Two tangents PA and PB are drawn from a point P at distance 10 cm from the centre O of a circle of radius 6 cm. Find the tangent length and the chord of contact AB. PA = √(100 − 36) = 8 cm. Let OP meet AB at M; M is the mid-point of AB and AM is the altitude of right triangle OAP on the hypotenuse OP, so AM = OA × AP/OP = 6 × 8/10 = 4.8 cm and AB = 9.6 cm.

Pattern 3: isosceles triangle with incircle. Prove that in an isosceles triangle with AB = AC, the incircle's point of contact D on BC is the mid-point of BC. BD = s − b and CD = s − c, and b = c, so BD = CD.

Pattern 4: tangents to concentric circles. Two concentric circles have radii 5 cm and 3 cm; find the length of the chord of the larger circle that touches the smaller. 2√(25 − 9) = 8 cm.

Pattern 5: circle inscribed in a right triangle. A circle is inscribed in a right triangle with legs 6 cm and 8 cm. Radius r = (6 + 8 − 10)/2 = 2 cm. The tangent lengths from the acute vertices are 6 − 2 = 4 cm and 8 − 2 = 6 cm, and their sum 10 cm is the hypotenuse, as it must be.

Pattern 6: angles. A tangent AB at a point A of a circle with centre O and a chord AC make ∠BAC = 60°; find ∠AOC and ∠OCA. OA ⊥ AB so ∠OAC = 30°; OA = OC so ∠OCA = 30°; hence ∠AOC = 120°. Note that the angle between the tangent and the chord, 60°, equals half the angle at the centre, 120°, subtended by the chord, and equals the angle in the alternate segment, a result seen in the Intermediate course.

Pattern 7: equal tangents around a circle. A circle touches the sides of a quadrilateral ABCD at P, Q, R, S with AP = 3, BQ = 4, CR = 5, DS = 6. Then AB = AP + PB = 3 + 4 = 7, BC = 4 + 5 = 9, CD = 5 + 6 = 11, DA = 6 + 3 = 9, and AB + CD = 18 = AD + BC.

Pattern 8: a rhombus-like figure. PQ is a chord of length 8 cm of a circle of radius 5 cm; the tangents at P and Q meet at T. Find TP. Let OT meet PQ at R; PR = 4 cm and OR = 3 cm. In right triangle OPT with PR the altitude to the hypotenuse OT, PR² = OR × RT gives 16 = 3 × RT, RT = 16/3 cm, and TP = √(PR² + RT²) = √(16 + 256/9) = √(400/9) = 20/3 cm.

Always draw the figure, join the centre to every point of contact to create right angles, and look for the right triangle that contains the unknown.

Though the syllabus concentrates on one circle, questions on two circles touching each other and on common tangents appear regularly, and they use only the theorems of this chapter.

Circles touching each other. Two circles touch each other if they have exactly one common point, the point of contact. If they touch externally, the distance between the centres equals the sum of the radii: d = r₁ + r₂. If they touch internally, d = r₁ − r₂. In either case the point of contact lies on the line joining the centres, and the common tangent at the point of contact is perpendicular to this line.

Number of common tangents. Two separate circles (d > r₁ + r₂) have four common tangents, two direct (the circles on the same side) and two transverse (the circles on opposite sides). Circles touching externally have three: two direct and the tangent at the point of contact. Intersecting circles (|r₁ − r₂| < d < r₁ + r₂) have two direct common tangents. Circles touching internally have one. If one circle lies inside the other without touching, there is none.

Length of a direct common tangent. For circles of radii r₁ ≥ r₂ with centres d apart, drop a perpendicular from the smaller centre to the larger radius drawn to the point of contact; it forms a right triangle with hypotenuse d and one leg r₁ − r₂, so the tangent length is √[d² − (r₁ − r₂)²]. For a transverse common tangent the leg is r₁ + r₂ and the length is √[d² − (r₁ + r₂)²].

Worked example. Two circles of radii 5 cm and 3 cm touch externally. Find the length of the direct common tangent. d = 8 cm; length = √(64 − 4) = √60 = 2√15 cm ≈ 7.75 cm.

Worked example. Prove that if two circles touch, the point of contact lies on the line joining the centres. The common tangent at the point of contact P is perpendicular to O₁P and to O₂P (Theorem 1 for each circle). Two perpendiculars to the same line at the same point P lie along one line, so O₁, P, O₂ are collinear.

Worked example. Two circles touch externally at P; a common tangent touches them at A and B, and the common tangent at P meets AB at T. Prove ∠APB = 90°. TA = TP and TB = TP (equal tangents from T to each circle), so T is equidistant from A, P, B, and P lies on the circle with diameter AB; hence ∠APB = 90°. Also TA = TB, so T is the mid-point of AB.

Worked example. Three circles of radii 1 cm, 2 cm, 3 cm touch each other externally in pairs. The triangle of centres has sides 3, 4, 5 cm and is right-angled, with area 6 cm².

These problems are excellent revision: every step is either tangent ⊥ radius or equal tangents from a point.