Circles | ICSE Class 10 Maths Notes

This note covers circle terminology, angles at the centre and circumference, cyclic quadrilaterals, tangents, touching circles, intersecting chords, secants, the alternate segment theorem, worked calculations, and constructions of tangents and circles associated with triangles and regular hexagons.

What terms and symbols describe a circle?

Definition: A circle is the collection of points in a plane at a fixed distance from a fixed point. The fixed point is its centre; the fixed distance is its radius.

A plane is a flat surface extending in all directions. A polygon is a closed plane figure formed by straight sides. A line segment joins two endpoints. A radius also means a segment from the centre to the circle. A chord joins two points on the circle; a diameter is a chord passing through its centre.

An arc is a part of the circle's boundary. Between endpoints that are not opposite ends of a diameter, the shorter arc is the minor arc and the longer arc is the major arc. A semicircle is half a circle, bounded by endpoints of a diameter.

A segment of a circle is the region bounded by a chord and one of its arcs. The smaller region is the minor segment and the larger region is the major segment. This use of segment differs from a straight line segment.

How should geometric notation be read?

Capital letters label points. In each new configuration, O denotes the centre when specified. AB denotes the segment joining points A and B, or its length in an equation. The symbol ∠ABC denotes the angle at B between BA and BC; ° means degrees.

The symbol = means equal to; + and − mean addition and subtraction; × means multiplication; a fraction bar or / means division. A superscript ² means the square of a quantity, and √ denotes the non-negative square root. The unit cm means centimetres.

Perpendicular lines meet at a right angle, measuring 90°. Parallel lines in a plane do not meet. A bisector divides an angle or segment into two equal parts. An angle is subtended by an arc when lines joining its endpoints to the vertex form that angle.

How are angles at the centre and circumference related?

Theorem: the angle at the centre is double

An arc subtends an angle at the centre equal to twice the angle it subtends at any point on the remaining part of the circle. The phrase remaining part identifies where the vertex must lie. Both angles must correspond to the same arc.

Let A and B be the arc's endpoints, O the centre and C a point on the remaining circumference, meaning the circle's boundary. The central angle corresponding to the arc AB that excludes C equals 2 × ∠ACB. Select that central angle, even if it is reflex.

A reflex angle is greater than 180° and less than 360°. The smaller angle between two radii does not automatically correspond to the arc under consideration. Identify the arc before doubling or halving an angle.

How does the proof use equal radii?

Join OA, OB and OC, and extend CO beyond O to a point D on the circle. A triangle with two equal sides is isosceles; its angles opposite those sides are equal. Thus OA = OC and OB = OC give two isosceles triangles.

1. In triangle AOC, OA = OC, so ∠OAC = ∠ACO. The exterior angle ∠AOD equals their sum, giving ∠AOD = 2 × ∠ACD.
2. Similarly, triangle BOC gives ∠BOD = 2 × ∠BCD. An exterior angle is formed by a side and the extension of an adjacent side.
3. If CD lies inside ∠ACB, add the two equalities. The corresponding central angles add to twice ∠ACB.
4. If CD lies outside ∠ACB, subtract the smaller equality from the larger. If CD coincides with a bounding side, use the single remaining equality. Each case gives the stated arc-angle relation.

Worked example 1. Points A, B and C lie on a circle with centre O, and ∠ABC = 45°. Find the central angle corresponding to arc AC that does not contain B.

Answer: That central angle is 2 × ∠ABC = 2 × 45° = 90°. Therefore the radii OA and OC are perpendicular. The angle at B and the central angle must intercept the same arc AC.

Why are angles in the same segment equal?

Theorem: angles in the same segment

Angles in the same segment of a circle are equal. Let AB be a chord and C and D be distinct points on the same arc between A and B. Then ∠ACB = ∠ADB because both angles stand on chord AB from the same side.

Check both the endpoints and the side of the chord. Sharing endpoints is insufficient if the vertices lie on opposite sides of the chord. In that configuration the angles are supplementary, meaning their sum is 180°, rather than necessarily equal.

Theorem: the angle in a semicircle

If AB is a diameter and C is any other point on the circle, ∠ACB = 90°. The diameter subtends a straight angle, measuring 180°, at the centre. The angle at C is half of that central angle, so it is a right angle.

This makes triangle ACB a right-angled triangle. AB is its hypotenuse, the side opposite the right angle. Pythagoras' theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides. Here AB² = AC² + BC².

Worked example 2. AB is a diameter of a circle, C lies on the circle, and AC = BC. Find ∠CAB.

Answer: ∠ACB = 90° because AB is a diameter. Equal sides AC and BC give equal angles at A and B. Since a triangle's angles total 180°, ∠CAB = (180° − 90°)/2 = 45°.

The equality AC = BC is essential to the final step. A diameter alone establishes the right angle at C; it does not make the other two angles equal. Keep each given condition attached to the conclusion it supports.

What angle properties do cyclic quadrilaterals have?

Definition: A cyclic quadrilateral is a four-sided polygon whose four vertices lie on one circle. A vertex is a corner where two sides meet.

Theorem: opposite angles are supplementary

Let A, B, C and D occur in that order around a circle. Then ∠DAB + ∠BCD = 180°, and ∠ABC + ∠CDA = 180°. These are opposite pairs: their members do not share a side of the quadrilateral.

To prove the first relation, consider the two arcs joining B and D. Angle DAB subtends the arc through C, and angle BCD subtends the arc through A. Their corresponding central angles together make a full turn of 360°.

Each angle at the circumference is half its corresponding central angle. Their sum is therefore half of 360°, or 180°. Applying the same reasoning to the two arcs joining A and C proves the other pair. The proof works without assuming equal sides or a special shape.

What happens when a side is extended?

Extend AB beyond B to E. The exterior angle CBE equals the opposite interior angle CDA. An interior angle lies inside the polygon. The word opposite matters: the exterior angle is supplementary to the adjacent interior angle ABC and equal to the interior angle at D.

Worked example 3. ABCD is a cyclic quadrilateral with vertices in order, AB is a diameter, and ∠ADC = 140°. Find ∠BAC.

Answer: Opposite angles give ∠ABC = 180° − 140° = 40°. The diameter gives ∠ACB = 90°. In triangle ABC, ∠BAC = 180° − 40° − 90° = 50°. Both the cyclic condition and the diameter are needed.

A useful converse, meaning a statement with the condition and conclusion reversed, is that a quadrilateral with a pair of supplementary opposite angles is cyclic. Do not assume a quadrilateral is cyclic merely because a circle can be sketched near its corners.

How does a tangent relate to the radius?

A tangent is a line meeting a circle at exactly one point, called its point of contact. A secant meets the circle at two points. A line that does not meet the circle is a non-intersecting line.

Theorem: radius and tangent are perpendicular

The tangent at any point of a circle is perpendicular to the radius through that point. If XY is tangent at P to a circle with centre O, then OP is perpendicular to XY. The line containing OP is also called the normal at P.

For the proof, take any point Q on XY other than P. It lies outside the circle, so OQ is greater than the radius OP. Thus OP is the shortest distance from O to the line. The shortest segment from a point to a line is perpendicular to it.

The circle has centre O and tangent XY through P. The figure shows OP meeting the tangent at a right angle, with Q elsewhere on the tangent and OQ drawn from the centre.

See Fig. 10.5 in your NCERT textbook

Worked example 4. PQ is tangent at P to a circle with centre O and radius 5 cm. Q lies on a line through O, and OQ = 12 cm. Find PQ.

Answer: OP is perpendicular to PQ. In right triangle OPQ, OQ is the hypotenuse. Therefore PQ² = OQ² − OP² = 12² − 5² = 119, so PQ = √119 cm. Use the positive square root for a length.

The right angle belongs at the contact point. It is not at the centre or at the external point. Joining the centre to the contact point supplies the right triangle needed for many length calculations.

Why are the two tangents from an external point equal?

Theorem: equal tangent lengths

Let P be outside a circle with centre O. Let PQ and PR touch it at Q and R. Then PQ = PR. Here tangent length means the finite segment between the external point and the point of contact, rather than the entire tangent line.

1. Join OP, OQ and OR. The angles OQP and ORP are right angles because radii are perpendicular to tangents at contact.
2. OQ = OR because both are radii of the same circle. OP is the common hypotenuse of the two right triangles.
3. The triangles OQP and ORP are congruent by the right angle, hypotenuse and side criterion. Congruent triangles have corresponding sides and angles equal.
4. Consequently PQ = PR. The corresponding angles at P are also equal, so OP bisects ∠QPR.

P lies outside the circle with centre O. The two tangents meet the circle at Q and R; the drawing includes OP and the radii OQ and OR.

See Fig. 10.7 in your NCERT textbook

How do the tangent and central angles combine?

In quadrilateral OQPR, the angles at Q and R are each 90°. Since a quadrilateral's interior angles total 360°, ∠QPR + ∠QOR = 180°. The angle between the two tangents is supplementary to the smaller central angle between the contact radii.

Worked example 5. TP and TQ are tangents at P and Q to a circle with centre O. The smaller ∠POQ is 110°. Find ∠PTQ.

Answer: The angles at P and Q in quadrilateral OPTQ are 90° each. Hence ∠PTQ = 360° − 90° − 90° − 110° = 70°. Equivalently, subtract the central angle from 180°.

The equality of tangent lengths requires a common external point. It does not compare arbitrary tangents drawn from different points. For proofs involving several tangent segments, group the equal segments by their shared external point before combining the equalities.

How can tangent lengths and related chords be calculated?

The first step is to identify a right triangle by joining a radius to a contact point. Distinguish the radius, the tangent segment and the segment from the centre to the external point. The last of these is the hypotenuse.

How do you recover an unknown radius?

Worked example 6. Q is 25 cm from the centre O of a circle. A tangent from Q touches the circle at T and has length QT = 24 cm. Find the radius OT.

Answer: OT is perpendicular to QT, so OT² = OQ² − QT² = 25² − 24² = 49. Therefore OT = 7 cm. The centre-to-external-point distance, 25 cm, is the hypotenuse; the tangent is a shorter side.

Worked example 7. A is 5 cm from the centre O of a circle. The tangent from A touches at T and has length AT = 4 cm. Find the radius.

Answer: Triangle OTA is right-angled at T. Hence OT² = OA² − AT² = 5² − 4² = 9, giving OT = 3 cm. The radius ends at the point of contact, not at the external point A.

What changes for concentric circles?

Concentric circles share a centre. If a chord of the larger circle touches the smaller circle, the common radius to the contact point is perpendicular to that chord. A perpendicular from a circle's centre to a chord bisects it.

Worked example 8. Two concentric circles have centre O and radii 5 cm and 3 cm. Chord AB of the larger circle touches the smaller circle at P. Find AB.

Answer: OP = 3 cm and OA = 5 cm. OP is perpendicular to AB, so AP = PB. In right triangle OPA, AP² = 5² − 3² = 16. Thus AP = 4 cm and AB = 2 × 4 = 8 cm.

How can a chord determine a tangent length?

Worked example 9. PQ is an 8 cm chord of a circle with centre O and radius 5 cm. Tangents at P and Q meet at T. Find TP.

Answer: Join OT and let R be its intersection with PQ. Equal tangent lengths and the angle bisector OT give PR = RQ = 4 cm and a right angle at R. Therefore OR = √(5² − 4²) = 3 cm.

Right triangles TRP and PRO are similar: their angles at R are right angles and ∠RTP = ∠RPO. Thus TP/PO = RP/RO = 4/3, giving TP = 20/3 cm.

Similar triangles have equal corresponding angles and proportional corresponding sides. The angle-angle criterion establishes similarity when two angles of one triangle equal two angles of another. Keep corresponding sides in the same order when writing their ratios.

T lies outside the circle with centre O. Tangents TP and TQ meet the ends of chord PQ, labelled 8 cm. OP is labelled 5 cm, and the dashed segment OT meets PQ at R.

See Fig. 10.10 in your NCERT textbook

What happens when two circles touch?

Theorem: the contact point lies on the line of centres

Two circles touch when they have exactly one common point. If their centres are O and P and their contact point is T, then O, P and T lie on one straight line. Points on one straight line are called collinear.

At T the circles have a common tangent, meaning one line tangent to both at their shared contact point. OT and PT are both perpendicular to this tangent. There is a unique perpendicular through T, so both centres lie on that line.

How do internal and external contact differ?

For external contact, the circles lie outside one another and T lies between their centres. Consequently the centre distance is the sum of the radii. In this configuration OP = OT + TP.

For internal contact, one circle lies inside the other and touches it at T. If the circle centred at O is larger, P lies between O and T. The centre distance is the difference of the radii: OP = OT − PT.

These are different positions, so inspect which circle contains the other before adding or subtracting lengths. A sketch should show the contact point on both boundaries and the line of centres passing through it. Two circles intersecting at two points do not satisfy this touching-circle hypothesis.

How do products of intersecting chord segments work?

Theorem: chords intersecting inside a circle

Suppose chords AB and CD intersect at P inside a circle. Each chord is divided into two pieces. Then PA × PB = PC × PD. Both factors on each side belong to the same original chord, and all lengths are measured from the intersection P.

Join AC and DB. Triangles APC and DPB have equal angles at P because vertically opposite angles, the opposite angles formed by two intersecting lines, are equal. Also ∠PAC = ∠PDB because they stand on chord BC in the same segment.

The triangles are therefore similar by the angle-angle criterion. Their corresponding sides give PA/PD = PC/PB. Multiplying by PD × PB gives PA × PB = PC × PD. This proves the product rule rather than merely inferring it from a drawing.

Theorem: secants intersecting outside a circle

Let P lie outside a circle. One secant meets it first at A and then at B; the other meets it first at C and then at D. The point orders are P, A, B and P, C, D. Again PA × PB = PC × PD.

Join AD and BC. Triangles PAD and PCB share the angle between the two secants at P. Also ∠PDA = ∠PBC, since these are the angles CDA and ABC standing on chord AC in the same segment.

Similarity gives PA/PC = PD/PB, which rearranges to the required equality. Here PA and PC are the external portions, while PB and PD are the whole distances from P to the farther intersections. AB and CD alone are not those whole distances.

The symbolic product looks the same in both configurations, but the positions of P differ. Write the point order before substitution. It prevents the common error of using just the portion inside the circle as the second factor.

What is the alternate segment theorem?

Theorem: the angle between a tangent and a chord

When a tangent touches a circle at A and AB is a chord, the angles between the tangent and the chord equal the angles in their corresponding alternate segments. For a chosen tangent-chord angle, the alternate segment lies on the other side of AB from that angle.

This theorem connects an angle outside the circle's interior with an angle at the circumference. There are two tangent rays from A, meaning the two half-lines starting at A. Select the ray before identifying the corresponding segment; the two tangent-chord angles are supplementary.

How can a diameter prove the theorem?

Draw diameter AD. If B differs from D, angle ABD is 90°. The angle between the tangent ray on the side of AD containing B and chord AB is 90° − ∠DAB. In triangle ABD, ∠ADB is also 90° − ∠DAB.

These two angles are equal. Every point C in the same segment of AB as D gives ∠ACB = ∠ADB. Hence the selected tangent-chord angle equals the angle in its alternate segment.

The other tangent-chord angle is supplementary to the first. Angles subtended by AB from opposite segments are also supplementary, so the other pair is equal. If AB itself is a diameter, both tangent-chord angles and both corresponding circumference angles are 90°.

Before using the theorem, identify three things: the point of contact, the chord through it, and the vertex in the alternate segment. A tangent at another point does not justify replacing the angle under consideration with a circumference angle.

How does a tangent combine with a secant?

Theorem: the tangent-secant product

Let P lie outside a circle, let PT touch it at T, and let a secant from P meet it first at A and then at B. Then PT² = PA × PB. The square is of the tangent segment from P to its contact point T.

The secant product uses the external portion PA and the whole distance PB. If the inside chord AB is supplied instead of PB, first write PB = PA + AB. The equivalent form is PT² = PA × (PA + AB).

Why is the tangent length squared?

1. Join TA and TB. Consider triangles PTA and PBT, keeping P, A and B in that order along the secant.
2. The angle at P is common to both triangles, since PA and PB lie along the same ray.
3. The alternate segment theorem gives ∠PTA = ∠PBT. Thus the triangles are similar by the angle-angle criterion.
4. Corresponding sides give PT/PB = PA/PT. Multiplication by PB × PT gives PT² = PA × PB.

The tangency condition is essential: PT must touch the circle at T. A line drawn from P to an arbitrary point of the circle cannot be substituted for the tangent. Similarly, the two distances on the right must come from one secant through P.

When finding PT, take the positive square root of the product because PT represents a length. When finding the inside chord AB, find the whole distance PB first and then subtract PA. State the requested segment, rather than stopping at an intermediate distance.

How are tangents and inscribed or circumscribed circles constructed?

How are two tangents drawn from an external point?

Let the given circle have centre O and let P be outside it. The construction uses the angle in a semicircle to locate the contact points precisely. A ruler and compasses create the required lines, midpoints and circles.

1. Join OP and construct its midpoint M, the point dividing OP into equal lengths.
2. Draw a circle with centre M and radius MO. Its diameter is OP.
3. Mark the intersections of this auxiliary circle with the given circle as T and U. An auxiliary circle is an additional circle used to perform the construction.
4. Join PT and PU. Since angles OTP and OUP stand on diameter OP, both are 90°. These perpendiculars to the contact radii are the required tangents.

The converse tangent property justifies the last step: a line perpendicular to a radius at its endpoint on the circle is tangent there. Any other point on that line is farther from the centre than the radius, so cannot lie on the circle.

How is a circle circumscribed about a triangle?

A circumcircle passes through all vertices of a polygon. For triangle ABC, construct the perpendicular bisectors of two sides, such as AB and AC. Their intersection O is the circumcentre. Draw the circle with centre O and radius OA.

A perpendicular bisector passes through a segment's midpoint at right angles. Every point on it is equally distant from the segment's endpoints. Thus OA = OB and OA = OC, so the circle passes through A, B and C. The triangle's vertices must be non-collinear.

How is a circle inscribed in a triangle?

An incircle lies inside a polygon and touches every side. Construct the internal angle bisectors at two vertices of triangle ABC. Let their intersection be I, the incentre. Drop a perpendicular from I to side BC, meeting it at D.

Draw the circle with centre I and radius ID, called the inradius. A point on an angle bisector has equal perpendicular distances from the angle's sides. Therefore I is equally distant from all three sides, and the circle touches each side. Its radius is a perpendicular distance, not a distance to a vertex.

How are circles associated with a regular hexagon?

A regular hexagon has six equal sides and six equal interior angles. For a given regular hexagon ABCDEF, join opposite vertices A to D and B to E. Their intersection O is its centre.

For the circumcircle, use centre O and radius OA. For the incircle, construct the perpendicular from O to side AB, meeting it at M, and use radius OM. Equal centre-to-vertex distances justify the first circle; equal centre-to-side distances justify the second.

To construct a regular hexagon in a given circle, keep the compasses set to the circle's radius and step off six successive chords around it. Join successive points. Each centre-and-side triangle is equilateral, meaning all three sides are equal, with central angle 60°.

To circumscribe a regular hexagon about a given circle, first mark those six equally spaced points, then construct the tangent at each point perpendicular to its radius. Intersections of consecutive tangents form the vertices. Retain construction arcs and label centres and contact points clearly.

Glossary

• Circle — The collection of points in a plane at a fixed distance from a fixed centre.
• Radius — A segment joining the centre to a point on the circle, or the length of that segment.
• Chord — A straight line segment whose two endpoints lie on the circumference of a circle.
• Diameter — A chord passing through the centre, with length equal to twice the radius.
• Arc — A portion of the boundary of a circle between two specified points.
• Segment — The region enclosed by a chord and one of the arcs joining its endpoints.
• Cyclic quadrilateral — A quadrilateral with all four of its vertices lying on the same circle.
• Tangent — A line meeting a circle at exactly one point, called the point of contact.
• Secant — A straight line that intersects a circle at two distinct points on its circumference.
• Concentric circles — Circles that have a common centre but have different radii.
• Alternate segment — The circle segment on the opposite side of a chord from the selected tangent-chord angle.
• Circumcircle — A circle passing through every vertex of the polygon with which it is associated.
• Incircle — A circle inside a polygon that touches each of its sides.
• Supplementary angles — Two angles whose measures add to one straight angle, or 180 degrees.

Common errors and misconceptions

• Misconception: Any two angles standing on one chord are equal. Correct: Their vertices must be in the same segment. Angles from opposite segments are supplementary.
• Misconception: The angle at the circumference is twice the central angle. Correct: The central angle corresponding to the same arc is twice the circumference angle.
• Misconception: Adjacent angles of every cyclic quadrilateral total 180°. Correct: The theorem concerns opposite angles; identify the opposite pair before subtracting.
• Misconception: Any radius is perpendicular to a tangent. Correct: Use the radius drawn to that tangent's point of contact.
• Misconception: Tangent segments drawn from different points must be equal. Correct: The equal-tangents theorem compares two tangents from the same external point.
• Misconception: An external secant product uses the external portion and the inside chord. Correct: It uses the external portion and the whole distance to the farther intersection.
• Misconception: A triangle's incircle is centred at the intersection of perpendicular bisectors. Correct: Internal angle bisectors locate the incentre; perpendicular bisectors locate the circumcentre.
• Misconception: The centre distance for touching circles is always the sum of their radii. Correct: Use the sum for external contact and the difference for internal contact.

Exam-style questions with model answers

Q1. A, B and C lie on a circle with centre O and ∠ABC = 45°. Find the central angle corresponding to arc AC not containing B, and state the relation between OA and OC. [2 marks]

1. The central angle subtending the same arc is double the circumference angle, so ∠AOC = 2 × 45° = 90°.
2. Since the angle between OA and OC is 90°, the two radii are perpendicular to one another.

Q2. ABCD is a cyclic quadrilateral with vertices in order. AB is a diameter and ∠ADC = 140°. Find ∠BAC, giving reasons. [3 marks]

1. Opposite angles of cyclic quadrilateral ABCD are supplementary. Therefore ∠ABC = 180° − ∠ADC = 180° − 140° = 40°.
2. Because AB is a diameter and C lies on the circle, angle ACB is an angle in a semicircle. Thus ∠ACB = 90°.
3. The angles of triangle ABC total 180°, so ∠BAC = 180° − 40° − 90° = 50°.

Q3. Q lies 25 cm from the centre O of a circle. A tangent QT touches the circle at T and is 24 cm long. Find the radius, showing the geometric reason and calculation. [3 marks]

1. Join OT. The radius to the point of contact is perpendicular to the tangent, so triangle OTQ is right-angled at T.
2. OQ is the hypotenuse. Pythagoras' theorem gives OT² = OQ² − QT² = 25² − 24² = 625 − 576 = 49.
3. The radius is the positive length OT = √49 = 7 cm. A negative square root cannot represent the radius.

Q4. Two concentric circles have centre O and radii 5 cm and 3 cm. Chord AB of the larger circle touches the smaller circle at P. Find AB with reasons. [4 marks]

1. Join OP and OA. Since AB touches the smaller circle at P, its radius OP is perpendicular to AB and OP = 3 cm.
2. For the larger circle, a perpendicular from its centre bisects a chord. Therefore AP = PB, while OA = 5 cm.
3. In right triangle OPA, AP² = OA² − OP² = 5² − 3² = 16. Hence AP = 4 cm.
4. The complete chord has both equal halves: AB = AP + PB = 4 + 4 = 8 cm.

Q5. P lies outside a circle with centre O. PQ and PR are tangents touching the circle at Q and R. Prove that PQ = PR and that OP bisects ∠QPR. [5 marks]

1. Join OQ, OR and OP to form triangles OQP and ORP. These joins connect the centre to both contact points and the common external point.
2. Each contact radius is perpendicular to its tangent. Consequently ∠OQP and ∠ORP are both right angles, with OP the hypotenuse in each triangle.
3. OQ = OR because they are radii of the same circle. OP is a common side and therefore has the same length in both triangles.
4. The triangles are congruent by the right angle, hypotenuse and side criterion. Corresponding tangent sides are therefore equal, giving PQ = PR.
5. The corresponding angles at P are equal as well: ∠QPO = ∠OPR. Thus OP divides ∠QPR into two equal angles and is its bisector.

Q6. A circle has centre O and P is a given external point. Describe the construction of the two tangents from P and justify why the constructed lines are tangents. [5 marks]

1. Join OP and construct its perpendicular bisector to locate midpoint M. Thus MO and MP are equal lengths along the same straight segment.
2. Draw the auxiliary circle with centre M and radius MO. It passes through both O and P, so OP is its diameter.
3. Label its two intersections with the given circle T and U. Join PT and PU, and join the contact radii OT and OU.
4. Angles OTP and OUP stand on diameter OP of the auxiliary circle. The angle in a semicircle is a right angle, so each measures 90°.
5. PT and PU are therefore perpendicular to the original circle's radii at their endpoints T and U. By the converse tangent property, they are the required tangents.

Q7. From an external point P, two secants meet a circle in the orders P, A, B and P, C, D. Prove PA × PB = PC × PD. [4 marks]

1. Join AD and BC to form triangles PAD and PCB. Their angles at P are equal because both are formed by the same two secant rays.
2. Since DP passes through C and BP passes through A, ∠PDA and ∠PBC stand on chord AC in the same segment and are equal.
3. The triangles are similar by the angle-angle criterion. Corresponding sides therefore satisfy PA/PC = PD/PB.
4. Multiplying by PC × PB gives PA × PB = PC × PD, with each product using the external portion and the whole secant distance.

Q8. PT is tangent at T to a circle. A secant from the same external point P meets the circle first at A and then at B. Prove PT² = PA × PB. [4 marks]

1. Join TA and TB. Triangles PTA and PBT have the same angle at P because A and B lie on the same secant ray from P.
2. The alternate segment theorem gives ∠PTA = ∠PBT: the angle between tangent PT and chord TA matches the corresponding circumference angle.
3. The triangles are similar by the angle-angle criterion. In the corresponding side order, PT/PB = PA/PT.
4. Multiply by PB × PT to obtain PT² = PA × PB. PB is the complete distance to B, so PB = PA + AB.

Key takeaways

• The central angle is twice the circumference angle subtended by the same arc; identify the arc before choosing the central angle.
• Angles in the same segment are equal, and an angle standing on a diameter is a right angle.
• Opposite angles of a cyclic quadrilateral are supplementary; its exterior angle equals the opposite interior angle.
• A tangent is perpendicular to the radius through its contact point, creating a right triangle for length calculations.
• Two tangents from one external point have equal lengths, and the line to the centre bisects their angle.
• For external secants, multiply the external portion by the whole distance; with a tangent, this product equals its squared length.
• The alternate segment theorem equates a tangent-chord angle with the circumference angle in the corresponding opposite segment.
• Perpendicular bisectors locate a triangle's circumcentre, while internal angle bisectors locate its incentre and perpendicular distances determine its inradius.

Test yourself

Which arc must be used when comparing a circumference angle with a central angle?

Use the arc between the angle's endpoints that does not contain its circumference vertex.

Why does a diameter produce a right angle at another point on the circle?

The corresponding central angle is 180°, and the circumference angle is half of it, giving 90°.

What distinguishes a cyclic quadrilateral from a quadrilateral circumscribing a circle?

A cyclic quadrilateral has its vertices on the circle; a circumscribing quadrilateral has its sides tangent to the circle.

Where is the right angle in a triangle formed by a tangent, contact radius and centre-to-external-point segment?

It is at the contact point, where the radius meets the tangent.

A secant meets a circle in the order P, A, B, with P outside. How is PB related to PA and AB?

The whole distance is PB = PA + AB because A lies between P and B.

For two externally touching circles, how is the centre distance found?

Add their radii because the contact point lies between the two centres.

Which circle is used to construct tangents from external point P to a circle with centre O?

Use the auxiliary circle with OP as diameter; its intersections locate the required contact points.

How is a triangle's incircle radius obtained once its incentre is known?

Drop a perpendicular from the incentre to any side and use that perpendicular distance as the radius.