ICSE Class 10 Maths Tangents Solutions | ICSE Board

Written by Greya LakshmiPublished on 19 May 2026Updated on 18 June 2026

ICSE Class 10 Maths Chapter 18 Tangents and Intersecting Chords

ICSE Class 10 Maths Chapter 18, Tangents and Intersecting Chords, is about using circle theorems to find lengths and prove angle or tangent relations. The main ideas are: a radius is perpendicular to the tangent at the point of contact, tangents from the same external point are equal, intersecting chords satisfy a product rule, and a tangent-secant pair also satisfies a product rule.

Use this page in this order: revise the theorem list, study the worked Selina/Concise textbook-style solutions, check the common mistakes, and then use the quick answer index for last-minute revision. The solutions use MathJax so that each formula is readable on the page.

Key theorems used in Chapter 18

Before solving the questions, revise the theorem that matches the diagram. Most errors in this chapter happen because students use the right theorem on the wrong line segment.

Situation in the figure	The result to use	How it is used
Tangent at $T$ and radius $OT$	$OT \perp PT$	Creates a right triangle, so Pythagoras can be used.
Two tangents from the same external point $P$	$PA = PB$	Turns tangent lengths into equal parts.
Two chords $AB$ and $CD$ intersect at $P$ inside a circle	$PA \times PB = PC \times PD$	Finds an unknown part of a chord.
One tangent $PT$ and one secant $PAB$ from an external point	$PT^2 = PA \times PB$	Here $PB$ is the whole secant from $P$ to the far point.
Two secants $PAB$ and $PCD$ from an external point	$PA \times PB = PC \times PD$	Each product uses the external segment and the whole secant.
Tangential quadrilateral $ABCD$	$AB + CD = BC + AD$	Comes from equal tangents from each vertex.

Concept snapshot: which theorem should you choose?

Think of every tangent or secant question as a line-segment matching task. If the line only touches the circle once, use the tangent rule. If two chords cross inside the circle, use the intersecting-chords product. If the point is outside and a line cuts the circle twice, use the secant product. The diagram tells the theorem before the numbers do.

Exercise 18(A): tangents and touching circles

The first set of questions is based mainly on equal tangents, radius perpendicular to tangent, tangential quadrilaterals and externally or internally touching circles. The solutions below follow the standard Selina Concise Mathematics Class 10 treatment of this chapter.

Question 1(a): Find $PA$ when $PA$ , $PB$ and $QR$ are tangents

Step 1: Let $QR$ touch the circle at $D$ . From the same external point, tangent lengths are equal.

Step 2: So, $PA = PB$ , $QD = QB$ , and $RD = RA$ .

Step 3: The perimeter of $\triangle PQR$ is given as $18\text{ cm}$ .

$PQ + QR + PR = 18$

$PQ + QD + RD + PR = 18$

Step 4: Replace $QD$ by $QB$ and $RD$ by $RA$ .

$PQ + QB + RA + PR = 18$

$PB + PA = 18$

Step 5: Since $PA = PB$ , let each be $x$ .

$x + x = 18 \Rightarrow 2x = 18 \Rightarrow x = 9$

Final answer: $PA = 9\text{ cm}$ .

Question 1(b): Chord of the outer circle touching the inner circle

Step 1: In the usual concentric-circles figure, $AB$ is a chord of the outer circle and touches the inner circle at $P$ . Given $OA = 10\text{ cm}$ and $OP = 6\text{ cm}$ .

Step 2: A radius drawn to the point of contact is perpendicular to the tangent, so $OP \perp AB$ .

Step 3: In right-angled $\triangle OAP$ , use Pythagoras.

$OA^2 = OP^2 + AP^2$

$10^2 = 6^2 + AP^2$

$AP^2 = 100 - 36 = 64$

$AP = 8\text{ cm}$

Step 4: The perpendicular from the centre to a chord bisects the chord. Hence $AP = PB = 8\text{ cm}$ .

$AB = AP + PB = 8 + 8 = 16\text{ cm}$

Final answer: $AB = 16\text{ cm}$ .

Question 1(c): Three touching circles and the length $AB$

Step 1: In the given type of figure, the distances between the centres are formed by adding the radii of the externally touching circles.

Step 2: From the given data, take $AC = 9 + 6 = 15\text{ cm}$ and $BC = 2 + 6 = 8\text{ cm}$ .

Step 3: If $\angle ACB = 90^\circ$ , apply Pythagoras in $\triangle ACB$ .

$AB^2 = AC^2 + BC^2$

$AB^2 = 15^2 + 8^2 = 225 + 64 = 289$

$AB = \sqrt{289} = 17\text{ cm}$

Final answer: $AB = 17\text{ cm}$ .

Question 1(d): Find $\angle B$ when $BC$ is a tangent

Step 1: In the standard figure, $OD$ and $OA$ are radii, so $OD = OA$ .

Step 2: Therefore $\triangle OAD$ is isosceles. Let $\angle ODA = \angle OAD = x$ .

Step 3: Given $\angle DOC = 100^\circ$ . This exterior angle is equal to the sum of the two opposite interior angles of $\triangle OAD$ .

$100^\circ = x + x$

$2x = 100^\circ \Rightarrow x = 50^\circ$

Step 4: Since $BC$ is tangent at $C$ , the radius to $C$ is perpendicular to the tangent, so $\angle C = 90^\circ$ .

Step 5: In $\triangle ABC$ , use angle sum.

$\angle B + 90^\circ + 50^\circ = 180^\circ$

$\angle B = 40^\circ$

Final answer: $\angle B = 40^\circ$ .

Question 1(e): Find $\angle ACB$ when $PA$ and $PB$ are tangents

Step 1: Join $OA$ and $OB$ . A radius is perpendicular to the tangent at the point of contact.

$\angle OAP = 90^\circ,\quad \angle OBP = 90^\circ$

Step 2: Given $\angle APB = 70^\circ$ . In quadrilateral $OAPB$ , the angle sum is $360^\circ$ .

$\angle AOB + 90^\circ + 70^\circ + 90^\circ = 360^\circ$

$\angle AOB = 110^\circ$

Step 3: The angle at the centre is twice the angle at the circumference standing on the same arc.

$\angle AOB = 2\angle ACB$

$\angle ACB = \frac{110^\circ}{2} = 55^\circ$

Final answer: $\angle ACB = 55^\circ$ .

Question 2: Length of a tangent from a point $10\text{ cm}$ from the centre

Step 1: Let $O$ be the centre, $T$ the point of contact and $P$ the external point. Given $OT = 8\text{ cm}$ and $OP = 10\text{ cm}$ .

Step 2: Since $OT$ is perpendicular to tangent $PT$ , $\triangle OTP$ is right-angled at $T$ .

$OP^2 = OT^2 + PT^2$

$10^2 = 8^2 + PT^2$

$PT^2 = 100 - 64 = 36$

$PT = 6\text{ cm}$

Final answer: The tangent length is $6\text{ cm}$ .

Question 3: Find the radius when $AB = 15\text{ cm}$ and $AC = 7.5\text{ cm}$

Step 1: Let the radius be $r\text{ cm}$ . Then $OB = r\text{ cm}$ and, from the figure type, $AO = AC + CO = 7.5 + r$ .

Step 2: Since $AB$ is tangent at $B$ , $OB \perp AB$ . So $\triangle AOB$ is right-angled at $B$ .

$AO^2 = AB^2 + OB^2$

$(7.5 + r)^2 = 15^2 + r^2$

Step 3: Expand and simplify.

$56.25 + 15r + r^2 = 225 + r^2$

$15r = 168.75$

$r = \frac{168.75}{15} = 11.25$

Final answer: The radius is $11.25\text{ cm}$ .

Question 4: Chord of outer circle touching inner circle

Step 1: The outer radius is $5\text{ cm}$ and the inner radius is $3\text{ cm}$ . The chord of the outer circle touches the inner circle, so the perpendicular distance from the common centre to the chord is $3\text{ cm}$ .

Step 2: Let half the chord be $x$ . In the right triangle formed,

$5^2 = 3^2 + x^2$

$x^2 = 25 - 9 = 16$

$x = 4\text{ cm}$

Step 3: The full chord is twice its half.

$2x = 2 \times 4 = 8\text{ cm}$

Final answer: The chord length is $8\text{ cm}$ .

Question 5: Radii of three externally touching circles

Step 1: Let the centres be $A$ , $B$ , and $C$ , and let their radii be $r_1$ , $r_2$ , and $r_3$ .

Step 2: When two circles touch externally, the distance between their centres equals the sum of their radii.

$r_1 + r_2 = 6 \quad \text{...(1)}$

$r_1 + r_3 = 8 \quad \text{...(2)}$

$r_2 + r_3 = 9 \quad \text{...(3)}$

Step 3: Add the three equations.

$2(r_1 + r_2 + r_3) = 6 + 8 + 9 = 23$

$r_1 + r_2 + r_3 = \frac{23}{2} = 11.5$

Step 4: Subtract equation $(3)$ from the total.

$r_1 = 11.5 - 9 = 2.5\text{ cm}$

Step 5: Subtract equation $(2)$ and equation $(1)$ similarly.

$r_2 = 11.5 - 8 = 3.5\text{ cm}$

$r_3 = 11.5 - 6 = 5.5\text{ cm}$

Final answer: The radii are $2.5\text{ cm}$ , $3.5\text{ cm}$ , and $5.5\text{ cm}$ .

Question 6: Prove $AB + CD = BC + AD$ for a tangential quadrilateral

Step 1: Let the circle touch $AB$ , $BC$ , $CD$ , and $DA$ at $P$ , $Q$ , $R$ , and $S$ , respectively.

Step 2: Equal tangents from the same external point give:

$AP = AS,\quad BP = BQ,\quad CQ = CR,\quad DR = DS$

Step 3: Add the equal parts in a way that forms full sides.

$(AP + BP) + (CR + DR) = (BQ + CQ) + (AS + DS)$

Step 4: Replace each pair by the corresponding side.

$AB + CD = BC + AD$

Hence proved.

Question 7: If a parallelogram touches a circle, prove that it is a rhombus

Step 1: Since the sides of parallelogram $ABCD$ touch a circle, it is a tangential quadrilateral.

Step 2: From the result proved for a tangential quadrilateral,

$AB + CD = BC + AD$

Step 3: In a parallelogram, opposite sides are equal.

$AB = CD,\quad BC = AD$

Step 4: Substitute these in the tangential quadrilateral relation.

$AB + AB = BC + BC$

$2AB = 2BC \Rightarrow AB = BC$

Step 5: Since adjacent sides of the parallelogram are equal, all four sides are equal.

Hence proved: The parallelogram is a rhombus.

Question 8: Prove $AP + BQ + CR = BP + CQ + AR$ and relate it to the perimeter

Step 1: In a triangle with an incircle touching $AB$ , $BC$ , and $CA$ at $P$ , $Q$ , and $R$ , equal tangents give:

$AP = AR,\quad BP = BQ,\quad CQ = CR$

Step 2: Add the left sides and right sides.

$AP + BQ + CR = AR + BP + CQ$

$AP + BQ + CR = BP + CQ + AR$

Step 3: Add $AP + BQ + CR$ to both sides.

$2(AP + BQ + CR) = (AP + BP) + (BQ + CQ) + (AR + CR)$

Step 4: Replace the sums by triangle sides.

$2(AP + BQ + CR) = AB + BC + CA$

$AP + BQ + CR = \frac{1}{2}(AB + BC + CA)$

Hence proved: $AP + BQ + CR = \dfrac{1}{2}$ of the perimeter of $\triangle ABC$ .

Question 9: If $AB = AC$ , prove $BQ = CQ$

Step 1: Use the same incircle notation: the circle touches $AB$ , $BC$ , and $CA$ at $P$ , $Q$ , and $R$ .

Step 2: Equal tangents give:

$AP = AR,\quad BP = BQ,\quad CQ = CR$

Step 3: Express the equal sides of the triangle using tangent parts.

$AB = AP + BP$

$AC = AR + CR$

Step 4: Given $AB = AC$ , substitute.

$AP + BP = AR + CR$

Step 5: Since $AP = AR$ , cancel equal parts.

$BP = CR$

Step 6: Replace $BP$ by $BQ$ and $CR$ by $CQ$ .

$BQ = CQ$

Hence proved.

Question 10: Distance between centres when two circles touch

Step 1: The radii are $6.3\text{ cm}$ and $3.6\text{ cm}$ .

Step 2: If the circles touch externally, the distance between centres is the sum of radii.

$d = 6.3 + 3.6 = 9.9\text{ cm}$

Step 3: If the circles touch internally, the distance between centres is the difference of radii.

$d = 6.3 - 3.6 = 2.7\text{ cm}$

Final answer: External touching: $9.9\text{ cm}$ . Internal touching: $2.7\text{ cm}$ .

Question 11: From an external point $P$ , tangents $PA$ and $PB$ are drawn

Step 1: Join $OA$ , $OB$ , and $OP$ , where $O$ is the centre of the circle.

Step 2: In $\triangle AOP$ and $\triangle BOP$ ,

$OA = OB \quad \text{(radii)}$

$PA = PB \quad \text{(equal tangents from } P\text{)}$

$OP = OP \quad \text{(common side)}$

Step 3: Therefore $\triangle AOP \cong \triangle BOP$ by SSS congruence.

$\angle AOP = \angle BOP$

Step 4: Let $OP$ meet chord $AB$ at $M$ . In $\triangle OAM$ and $\triangle OBM$ ,

$OA = OB,\quad \angle AOM = \angle BOM,\quad OM = OM$

Step 5: So $\triangle OAM \cong \triangle OBM$ by SAS congruence.

$AM = MB$

Step 6: Also, $\angle OMA = \angle OMB$ . Since these two angles form a straight line,

$\angle OMA + \angle OMB = 180^\circ$

$\angle OMA = \angle OMB = 90^\circ$

Hence proved: $\angle AOP = \angle BOP$ , and $OP$ is the perpendicular bisector of chord $AB$ .

Question 12: Two circles touch externally and $AB$ is a direct common tangent

Step 1: Let the common tangent at the point of contact $P$ meet $AB$ at $T$ .

Step 2: From point $T$ , the two tangents to the first circle are equal.

$TA = TP$

Step 3: From point $T$ , the two tangents to the second circle are equal.

$TB = TP$

Step 4: Hence $TA = TB$ . Therefore, $T$ bisects $AB$ .

Step 5: Since $TA = TP$ , $\triangle ATP$ is isosceles. Let $\angle TAP = \angle APT = \alpha$ .

Step 6: Since $TB = TP$ , $\triangle BTP$ is isosceles. Let $\angle TBP = \angle BPT = \beta$ .

Step 7: In $\triangle APB$ , the angles at $A$ and $B$ are $\alpha$ and $\beta$ , and $\angle APB = \alpha + \beta$ .

$\alpha + \beta + (\alpha + \beta) = 180^\circ$

$2(\alpha + \beta) = 180^\circ$

$\alpha + \beta = 90^\circ$

Hence proved: $\angle APB = 90^\circ$ .

Exercise 18(B): intersecting chords, secants and tangent-secant

Intersecting chords and secants are product-theorem questions. The important point is to multiply the correct pair of segments: for two chords inside the circle, use the two parts of each chord; for a secant from outside, use the external part and the whole secant.

Worked example 1: Intersecting chords inside a circle

Step 1: Suppose chords $AB$ and $CD$ intersect at $P$ inside a circle. Given $AP = 4\text{ cm}$ , $PB = 9\text{ cm}$ , and $CP = 6\text{ cm}$ . Find $DP$ .

Step 2: Use the intersecting-chords theorem.

$AP \times PB = CP \times DP$

$4 \times 9 = 6 \times DP$

$36 = 6DP$

$DP = 6\text{ cm}$

Final answer: $DP = 6\text{ cm}$ .

Worked example 2: Tangent-secant theorem

Step 1: From an external point $P$ , $PT$ is a tangent and $PAB$ is a secant. Given $PT = 12\text{ cm}$ and $PA = 8\text{ cm}$ . Find $PB$ and $AB$ .

Step 2: Use the tangent-secant theorem.

$PT^2 = PA \times PB$

$12^2 = 8 \times PB$

$144 = 8PB$

$PB = 18\text{ cm}$

Step 3: Since $PB$ is the whole secant and $PA$ is the external part,

$AB = PB - PA = 18 - 8 = 10\text{ cm}$

Final answer: $PB = 18\text{ cm}$ and $AB = 10\text{ cm}$ .

Worked example 3: Two secants from the same external point

Step 1: From an external point $P$ , two secants $PAB$ and $PCD$ are drawn. Given $PA = 5\text{ cm}$ , $PB = 20\text{ cm}$ , and $PC = 8\text{ cm}$ . Find $PD$ and $CD$ .

Step 2: For two secants from the same external point, multiply external segment by whole secant.

$PA \times PB = PC \times PD$

$5 \times 20 = 8 \times PD$

$100 = 8PD$

$PD = 12.5\text{ cm}$

Step 3: Since $PD$ is the whole secant and $PC$ is the external part,

$CD = PD - PC = 12.5 - 8 = 4.5\text{ cm}$

Final answer: $PD = 12.5\text{ cm}$ and $CD = 4.5\text{ cm}$ .

Worked example 4: Choosing the correct theorem

Step 1: A tangent $PT$ and a secant $PAB$ are drawn from $P$ . Given $PA = 4\text{ cm}$ and $AB = 12\text{ cm}$ . Find $PT$ .

Step 2: First find the whole secant.

$PB = PA + AB = 4 + 12 = 16\text{ cm}$

Step 3: Now use the tangent-secant theorem.

$PT^2 = PA \times PB$

$PT^2 = 4 \times 16 = 64$

$PT = 8\text{ cm}$

Final answer: $PT = 8\text{ cm}$ .

Examiner’s mindset for circle-theorem proofs

In ICSE Class 10 Maths geometry, a proof is usually checked for three things: the correct theorem, correct use of the theorem on named segments or angles, and a clear final statement. For example, writing only $PA = PB$ is not enough in a proof unless you also show that $PA$ and $PB$ are tangents drawn from the same external point.

For length questions, write the theorem before substituting numbers. This avoids the common error of using the inside-chord formula $PA \times PB = PC \times PD$ on a tangent-secant figure where $PT^2 = PA \times PB$ is required.

Common mistakes in tangents and intersecting chords

Using chord theorem on a tangent: If the line touches the circle at exactly one point, use the tangent theorem, not the intersecting-chords theorem.
Forgetting the whole secant: In $PT^2 = PA \times PB$ , $PB$ is the whole secant from the external point to the far point of the circle. It is not only the part inside the circle.
Not proving perpendicularity: Before using Pythagoras with a tangent, state $OT \perp PT$ because the radius to the point of contact is perpendicular to the tangent.
Mixing external and internal touching of circles: Externally touching circles have centre distance equal to the sum of radii; internally touching circles have centre distance equal to the difference of radii.
Leaving proof conclusions incomplete: End with the exact statement required, such as $AB + CD = BC + AD$ or $OP$ is the perpendicular bisector of $AB$ .

Quick answer index

Section	Question / type	Answer or result
Exercise 18(A)	Question 1(a)	$PA = 9\text{ cm}$
Exercise 18(A)	Question 1(b)	$AB = 16\text{ cm}$
Exercise 18(A)	Question 1(c)	$AB = 17\text{ cm}$
Exercise 18(A)	Question 1(d)	$\angle B = 40^\circ$
Exercise 18(A)	Question 1(e)	$\angle ACB = 55^\circ$
Exercise 18(A)	Question 2	Tangent length $= 6\text{ cm}$
Exercise 18(A)	Question 3	Radius $= 11.25\text{ cm}$
Exercise 18(A)	Question 4	Chord length $= 8\text{ cm}$
Exercise 18(A)	Question 5	Radii $= 2.5\text{ cm}, 3.5\text{ cm}, 5.5\text{ cm}$
Exercise 18(A)	Question 6	$AB + CD = BC + AD$
Exercise 18(A)	Question 7	The parallelogram is a rhombus.
Exercise 18(A)	Question 8	$AP + BQ + CR = BP + CQ + AR = \frac{1}{2}$ perimeter of $\triangle ABC$
Exercise 18(A)	Question 9	$BQ = CQ$
Exercise 18(A)	Question 10	External: $9.9\text{ cm}$ ; internal: $2.7\text{ cm}$
Exercise 18(A)	Question 11	$\angle AOP = \angle BOP$ , and $OP$ is the perpendicular bisector of $AB$ .
Exercise 18(A)	Question 12	$T$ bisects $AB$ , and $\angle APB = 90^\circ$ .
Exercise 18(B)	Intersecting chords	$AP \times PB = CP \times DP$
Exercise 18(B)	Tangent-secant	$PT^2 = PA \times PB$
Exercise 18(B)	Two secants	$PA \times PB = PC \times PD$

For revision around this chapter, students may also use ICSE Class 10 solutions, Concise Mathematics Selina Class 10 chapter-wise solutions, and Selina Class 10 circles solutions. For board syllabus reference, use the CISCE official website. For overlapping circle concepts, students can compare with NCERT textbook resources.

Frequently Asked Questions

Which theorem is most important in ICSE Class 10 Maths Chapter 18?

The most used theorem is that tangents from the same external point are equal, written as $PA = PB$ . For intersecting-chords questions, the key theorem is $PA \times PB = PC \times PD$ .

How do I know whether to use

PT^2 = PA \times PB

Use $PT^2 = PA \times PB$ when $PT$ is a tangent from an external point $P$ and $PAB$ is a secant from the same point. Remember that $PB$ is the whole secant, not only the part inside the circle.

Why is the radius perpendicular to a tangent?

The radius drawn to the point of contact is perpendicular to the tangent because the tangent touches the circle at exactly one point. In solutions, this gives a right angle and often allows the use of Pythagoras.

What is the difference between external and internal touching of circles?

If two circles touch externally, the distance between their centres is the sum of their radii. If they touch internally, the distance between their centres is the difference of their radii.

How should I write proof answers in Concise Mathematics Selina Solutions Class 10 ICSE Chapter 18 Tangents and Intersecting Chords?

Start by naming the theorem, write equal segments or angle relations with correct labels, then combine them to reach the exact statement required. Do not write only the final equality without showing where it came from.

ICSE Class 10 Maths Tangents Solutions | ICSE Board

ICSE Class 10 Maths Chapter 18 Tangents and Intersecting Chords

Key theorems used in Chapter 18

Concept snapshot: which theorem should you choose?

Exercise 18(A): tangents and touching circles

Question 1(a): Find $PA$ when $PA$ , $PB$ and $QR$ are tangents

Question 1(b): Chord of the outer circle touching the inner circle

Question 1(c): Three touching circles and the length $AB$

Question 1(d): Find $\angle B$ when $BC$ is a tangent

Question 1(e): Find $\angle ACB$ when $PA$ and $PB$ are tangents

Question 2: Length of a tangent from a point $10\text{ cm}$ from the centre

Question 3: Find the radius when $AB = 15\text{ cm}$ and $AC = 7.5\text{ cm}$

Question 4: Chord of outer circle touching inner circle

Question 5: Radii of three externally touching circles

Question 6: Prove $AB + CD = BC + AD$ for a tangential quadrilateral

Question 7: If a parallelogram touches a circle, prove that it is a rhombus

Question 8: Prove $AP + BQ + CR = BP + CQ + AR$ and relate it to the perimeter

Question 9: If $AB = AC$ , prove $BQ = CQ$

Question 10: Distance between centres when two circles touch

Question 11: From an external point $P$ , tangents $PA$ and $PB$ are drawn

Question 12: Two circles touch externally and $AB$ is a direct common tangent

Exercise 18(B): intersecting chords, secants and tangent-secant

Worked example 1: Intersecting chords inside a circle

Worked example 2: Tangent-secant theorem

Worked example 3: Two secants from the same external point

Worked example 4: Choosing the correct theorem

Examiner’s mindset for circle-theorem proofs

Common mistakes in tangents and intersecting chords

Quick answer index

Frequently Asked Questions

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ICSE Class 10 Maths Chapter 18 Tangents and Intersecting Chords

Key theorems used in Chapter 18

Concept snapshot: which theorem should you choose?

Exercise 18(A): tangents and touching circles

Question 1(a): Find PA when PA, PB and QR are tangents

Question 1(b): Chord of the outer circle touching the inner circle

Question 1(c): Three touching circles and the length AB

Question 1(d): Find \angle B when BC is a tangent

Question 1(e): Find \angle ACB when PA and PB are tangents

Question 2: Length of a tangent from a point 10\text{ cm} from the centre

Question 3: Find the radius when AB = 15\text{ cm} and AC = 7.5\text{ cm}

Question 4: Chord of outer circle touching inner circle

Question 5: Radii of three externally touching circles

Question 6: Prove AB + CD = BC + AD for a tangential quadrilateral

Question 7: If a parallelogram touches a circle, prove that it is a rhombus

Question 8: Prove AP + BQ + CR = BP + CQ + AR and relate it to the perimeter

Question 9: If AB = AC, prove BQ = CQ

Question 10: Distance between centres when two circles touch

Question 11: From an external point P, tangents PA and PB are drawn

Question 12: Two circles touch externally and AB is a direct common tangent

Exercise 18(B): intersecting chords, secants and tangent-secant

Worked example 1: Intersecting chords inside a circle

Worked example 2: Tangent-secant theorem

Worked example 3: Two secants from the same external point

Worked example 4: Choosing the correct theorem

Examiner’s mindset for circle-theorem proofs

Common mistakes in tangents and intersecting chords

Quick answer index

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Frequently Asked Questions

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Question 1(a): Find $PA$ when $PA$ , $PB$ and $QR$ are tangents

Question 1(c): Three touching circles and the length $AB$

Question 1(d): Find $\angle B$ when $BC$ is a tangent

Question 1(e): Find $\angle ACB$ when $PA$ and $PB$ are tangents

Question 2: Length of a tangent from a point $10\text{ cm}$ from the centre

Question 3: Find the radius when $AB = 15\text{ cm}$ and $AC = 7.5\text{ cm}$

Question 6: Prove $AB + CD = BC + AD$ for a tangential quadrilateral

Question 8: Prove $AP + BQ + CR = BP + CQ + AR$ and relate it to the perimeter

Question 9: If $AB = AC$ , prove $BQ = CQ$

Question 11: From an external point $P$ , tangents $PA$ and $PB$ are drawn

Question 12: Two circles touch externally and $AB$ is a direct common tangent