Focus Keyword: NCERT Solutions Class 10 Math Chapter 10 Circles
Secondary Keywords & LSI: CBSE Class 10 Maths Circles solutions, NCERT Class 10 Maths Chapter 10 Exercise 10.2, Class 10 Circles theorems proofs, tangent to a circle Class 10 CBSE marking scheme, board exam preparation Class 10 Maths
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H1 Title: NCERT Solutions for Class 10 Math Chapter 10: Circles (Complete Step-by-Step Guide)
Navigating through the CBSE Class 10 Mathematics curriculum requires an in-depth understanding of planar geometry, the point of contact, tangent properties, and cyclic configurations. Chapter 10 of Class 10 Mathematics, “Circles”, forms the foundation of advanced coordinate geometry, trigonometry applications, and structural engineering mechanics. It investigates the geometrical relationship between a line and a circle; explores the perpendicularity of the tangent and radius at the point of contact; and details the fundamental theorem regarding the equality of tangent lengths drawn from an external point. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from in-text question sets and chapter-end exercises to complete theorem proofs, Solved Examples 1 to 3, and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms and practical protocols have been highlighted to ensure students secure maximum marks in their CBSE Board Examinations.
Chapter 10: Circles
Master Chapter Summary & Strategic Blueprint
The mathematical study of circles in Class 10 transitions from basic chord properties to the analytical study of tangents, secants, and circumscribed polygons. Under the rationalised CBSE curriculum, mastery of two foundational theorems forms the core requirement for solving all multi-step geometric proofs and numerical problems.
| Geometric Entity / Theorem | Formal Mathematical Statement | Core Algebraic / Geometric Relationship | Typical Board Error | CBSE Marks Weightage |
|---|---|---|---|---|
| Secant | A straight line intersecting a circle at two distinct points | Intersects at points $A$ and $B$ | Confusing a secant line with an interior chord | 1 Mark (Objective) |
| Tangent Line | A line that touches the circle at exactly one single point | Point of Contact $P$; common to line and circle | Assuming a tangent can cross inside the circle | 1 Mark (MCQ / Fill-in) |
| Theorem 10.1 | The tangent at any point is perpendicular to the radius through the point of contact | $OP \perp XY$ at point of contact $P$ | Omitting “shortest distance is perpendicular” proof | 3 Marks (Proof) |
| Theorem 10.2 | Lengths of tangents drawn from an external point to a circle are equal | $PQ = PR$ from external point $P$ | Forgetting to specify RHS congruency criteria | 3 to 5 Marks (Proof / CBQ) |
| Circumscribed Quadrilateral | A 4-sided polygon whose sides are tangent to an enclosed circle | $AB + CD = AD + BC$ | Misidentifying pairs of tangent segments | 3 to 4 Marks |
🧠 Examiner’s Secret: When proving geometric riders in Chapter 10, the CBSE marking scheme strictly allocates marks in four stages: (1) Given & To Prove ($0.5$ mark), (2) Accurate, labelled diagram ($0.5$ to $1$ mark), (3) Construction ($0.5$ mark, if applicable), and (4) Logical step-by-step Proof with mathematical justifications in brackets ($1.5$ to $2.5$ marks). Never skip the structural breakdown.
Foundational Geometric Theorems (Complete Theorem Vault)
Tangent to a Circle
A tangent to a circle is a straight line that intersects or touches the circle at only one single point.
The common point where the straight line touches the circumference of the circle is called the point of contact. At this precise location, the tangent remains entirely outside the circle without cutting across its interior.
Theorem 10.1: Radius-Tangent Perpendicularity Theorem
Theorem 10.1 states that the tangent at any point of a circle is perpendicular to the radius through the point of contact. O (Centre) P (Point of Contact) Q X Y
Given:
A circle $C(O, r)$ with centre $O$ and a tangent line $XY$ touching the circle at point $P$.
To Prove:
$$OP \perp XY$$
Construction:
Take any point $Q$ other than $P$ on the tangent line $XY$. Join the line segment $OQ$. Let $OQ$ intersect the circle at point $R$.
Proof:
- Point $Q$ must lie outside the circle.
(If $Q$ were to lie inside the circle, the line $XY$ would cut the circle at two points, becoming a secant rather than a tangent). - Because point $R$ lies on the circumference of the circle:
$$OR = OP = r \quad (\text{Radii of the same circle})$$ - From the geometric construction:
$$OQ = OR + RQ$$ - Since $RQ > 0$, it follows that:
$$OQ > OR$$ - Substituting $OR = OP$:
$$OQ > OP$$ - This inequality holds true for every single point on the line $XY$ except the point of contact $P$.
- Therefore, $OP$ is the shortest distance from the centre $O$ to any point on the line $XY$.
- By the fundamental geometric principle that the shortest distance between a point and a line is the perpendicular distance:
$$OP \perp XY$$
Hence Proved.
Theorem 10.2: Tangents from an External Point
Theorem 10.2 states that the lengths of tangents drawn from an external point to a circle are equal. O P Q R
Given:
A circle with centre $O$, an external point $P$, and two tangents $PQ$ and $PR$ drawn from $P$ touching the circle at $Q$ and $R$, respectively.
To Prove:
$$PQ = PR$$
Construction:
Join $OP$, $OQ$, and $OR$.
Proof:
- According to Theorem 10.1, the radius drawn to the point of contact is perpendicular to the tangent:
$$OQ \perp PQ \implies \angle OQP = 90^\circ$$
$$OR \perp PR \implies \angle ORP = 90^\circ$$ - Now consider the two right-angled triangles $\triangle OQP$ and $\triangle ORP$:
- $\angle OQP = \angle ORP = 90^\circ$ [From Theorem 10.1]
- $OP = OP$ [Common hypotenuse]
- $OQ = OR$ [Radii of the same circle]
- By the RHS (Right Angle-Hypotenuse-Side) congruence criterion:
$$\triangle OQP \cong \triangle ORP$$ - Therefore, by Corresponding Parts of Congruent Triangles (CPCTC):
$$PQ = PR$$
Hence Proved.
💡 Did You Know?: Theorem 10.2 can also be proved using the Pythagoras Theorem:
$$PQ = \sqrt{OP^2 – OQ^2}$$
$$PR = \sqrt{OP^2 – OR^2}$$
Because $OQ = OR = r$, it directly follows that $PQ = PR$.
[👉 Also Read: Class 10 Math Chapter 9 Some Applications of Trigonometry NCERT Solutions]
Step-by-Step Solutions: NCERT Class 10 Mathematics Chapter 10 Solved Examples
Example 1 (Page 148) [CBSE 2017, 2019, 2024 Set-1]
Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Answer: O A B P
Given Data:
Two concentric circles $C_1$ and $C_2$ with a common centre $O$. $AB$ is a chord of the larger circle $C_1$ touching the smaller circle $C_2$ at the point $P$.
To Prove:
$$AP = BP$$
Construction:
Join $OP$.
Step-by-Step Proof:
- The line $AB$ touches the smaller circle $C_2$ at the point $P$. Therefore, $AB$ acts as a tangent to $C_2$ at point $P$.
- By Theorem 10.1, the radius drawn through the point of contact is perpendicular to the tangent:
$$OP \perp AB$$ - Now, consider $AB$ as a chord of the larger circle $C_1$.
- By the standard circle theorem (the perpendicular drawn from the centre of a circle to a chord bisects the chord):
$$OP \perp AB \implies AP = BP$$
Conclusion:
Hence proved, the chord of the larger circle is bisected at the point of contact $P$.
Example 2 (Page 149) [CBSE 2016, 2018, 2020 Standard, 2023]
Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2\angle OPQ$.
Answer: O T P Q
Given Data:
A circle with centre $O$. From an external point $T$, two tangents $TP$ and $TQ$ touch the circle at points $P$ and $Q$.
To Prove:
$$\angle PTQ = 2\angle OPQ$$
Step-by-Step Proof:
- Let:
$$\angle PTQ = \theta$$ - By Theorem 10.2, the lengths of tangents drawn from an external point to a circle are equal:
$$TP = TQ$$ - Since $TP = TQ$, $\triangle TPQ$ is an isosceles triangle. Therefore, the angles opposite to these sides are equal:
$$\angle TPQ = \angle TQP$$ - In $\triangle TPQ$, the sum of interior angles is $180^\circ$:
$$\angle PTQ + \angle TPQ + \angle TQP = 180^\circ$$
$$\theta + 2\angle TPQ = 180^\circ$$
$$2\angle TPQ = 180^\circ – \theta$$
$$\angle TPQ = \frac{1}{2}(180^\circ – \theta) = 90^\circ – \frac{1}{2}\theta \quad \text{— (Equation 1)}$$ - By Theorem 10.1, the radius $OP$ is perpendicular to the tangent $TP$ at the point of contact:
$$\angle OPT = 90^\circ$$ - From the diagram, $\angle OPT$ can be expressed as:
$$\angle OPQ + \angle TPQ = 90^\circ$$
$$\angle OPQ = 90^\circ – \angle TPQ$$ - Substitute Equation 1 into this expression:
$$\angle OPQ = 90^\circ – \left(90^\circ – \frac{1}{2}\theta\right)$$
$$\angle OPQ = 90^\circ – 90^\circ + \frac{1}{2}\theta$$
$$\angle OPQ = \frac{1}{2}\theta$$ - Multiply both sides by $2$:
$$2\angle OPQ = \theta$$ - Since $\theta = \angle PTQ$:
$$\angle PTQ = 2\angle OPQ$$
Conclusion:
Hence Proved.
Example 3 (Page 150) [CBSE 2015, 2018, 2021 Term-2, 2024]
$PQ$ is a chord of length $8\text{ cm}$ of a circle of radius $5\text{ cm}$. The tangents at $P$ and $Q$ intersect at a point $T$ (see Fig. 10.10). Find the length $TP$.
Answer: O T P Q R
Given Data:
- Radius of the circle, $OP = 5\text{ cm}$.
- Length of chord, $PQ = 8\text{ cm}$.
- $TP$ and $TQ$ are tangents meeting at external point $T$.
To Find:
Length of tangent $TP$.
Step 1: Symmetry and Bisector Analysis
Join $OT$. Let $OT$ intersect the chord $PQ$ at point $R$.
- Since $TP = TQ$ (Theorem 10.2), $\triangle TPQ$ is isosceles.
- $OT$ is the angle bisector of $\angle PTQ$.
- In an isosceles triangle, the angle bisector of the vertex angle is perpendicular to the base and bisects it.
- Therefore:
$$OT \perp PQ \quad \text{and} \quad PR = RQ = \frac{PQ}{2} = \frac{8}{2} = 4\text{ cm}$$
Step 2: Calculate $OR$ using Pythagoras Theorem in $\triangle PRO$
In right-angled triangle $\triangle PRO$, $\angle PRO = 90^\circ$:
$$OP^2 = PR^2 + OR^2$$
$$5^2 = 4^2 + OR^2$$
$$25 = 16 + OR^2 \implies OR^2 = 9 \implies OR = 3\text{ cm}$$
Step 3: Determine $TP$ using Triangle Similarity
Consider right-angled $\triangle OPT$ ($\angle OPT = 90^\circ$ by Theorem 10.1) and right-angled $\triangle PRO$ ($\angle PRO = 90^\circ$):
- $\angle OPT = \angle PRO = 90^\circ$
- $\angle POT = \angle POR$ (Common angle)
- By AA Similarity criterion:
$$\triangle OPT \sim \triangle RPO$$
Set up the ratio of corresponding sides:
$$\frac{TP}{PR} = \frac{OP}{OR}$$
Substitute the known lengths:
$$\frac{TP}{4} = \frac{5}{3}$$
$$TP = \frac{4 \times 5}{3} = \frac{20}{3}\text{ cm}$$
Final Answer:
The length of the tangent $TP$ is $\frac{20}{3}\text{ cm}$ (or $6\frac{2}{3}\text{ cm}$).
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 10.1
Question 1 (Page 150) [CBSE 2013, 2019 Objective]
How many tangents can a circle have?
Answer:
A circle can have infinitely many tangents.
Mathematical Justification:
A circle is defined as a locus of an infinite number of points situated at an equal distance from the centre. By Theorem 10.1, a unique tangent can be drawn at each point on the circumference. Because there are infinitely many points on the circle, an infinite number of tangents can be constructed.
Question 2 (Page 150) [CBSE 2014, 2020 Objective]
Fill in the blanks:
(i) A tangent to a circle intersects it in _______ point(s).
(ii) A line intersecting a circle in two points is called a _______.
(iii) A circle can have _______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _______.
Answer:
(i) one
(A tangent touches the circle at exactly one distinct location).
(ii) secant
(A straight line cutting through the boundary at two distinct points is defined as a secant).
(iii) two
(Parallel tangents can only be drawn at the exact opposite ends of a diameter; hence, a circle can have at most two parallel tangents for any given orientation).
(iv) point of contact
(The single shared point between the tangent line and the perimeter is the point of contact).
Question 3 (Page 150) [CBSE 2018, 2023 Set-1]
A tangent $PQ$ at a point $P$ of a circle of radius $5\text{ cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12\text{ cm}$. Length $PQ$ is:
(A) $12\text{ cm}$
(B) $13\text{ cm}$
(C) $8.5\text{ cm}$
(D) $\sqrt{119}\text{ cm}$
Answer:
The correct option is (D) $\sqrt{119}\text{ cm}$.
Step-by-Step Mathematical Evaluation:
- Identify the given parameters:
- Radius of the circle, $OP = 5\text{ cm}$.
- Distance from centre $O$ to point $Q$, $OQ = 12\text{ cm}$.
- By Theorem 10.1, the tangent $PQ$ is perpendicular to the radius $OP$ at the point of contact $P$:
$$\angle OPQ = 90^\circ$$ - Apply Pythagoras Theorem in right-angled triangle $\triangle OPQ$:
$$OQ^2 = OP^2 + PQ^2$$
$$12^2 = 5^2 + PQ^2$$
$$144 = 25 + PQ^2$$
$$PQ^2 = 144 – 25 = 119$$
$$PQ = \sqrt{119}\text{ cm}$$
Final Answer:
The length $PQ$ is $\sqrt{119}\text{ cm}$.
Question 4 (Page 150) [CBSE Practical Geometry Standard]
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer: Line l (Given) Line m (Tangent) Line n (Secant) Point of Contact
Procedure of Construction:
- Draw a circle with centre $O$ of arbitrary radius.
- Draw a given straight line $l$ outside the circle.
- Draw a perpendicular line from centre $O$ to line $l$, meeting line $l$ at a point $M$.
- Through the intersection point of this perpendicular line with the circle’s boundary, draw line $m$ perpendicular to $OM$. Line $m$ touches the circle at exactly one point, making it a tangent parallel to line $l$.
- Select an interior point along the radius and draw a straight line $n$ perpendicular to $OM$. Line $n$ cuts the circle at two distinct points, making it a secant parallel to line $l$.
[👉 Also Read: Class 10 Math Chapter 11 Areas Related to Circles NCERT Solutions]
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 10.2
Question 1 (Page 153) [CBSE 2017, 2020 Objective]
From a point $Q$, the length of the tangent to a circle is $24\text{ cm}$ and the distance of $Q$ from the centre is $25\text{ cm}$. The radius of the circle is:
(A) $7\text{ cm}$
(B) $12\text{ cm}$
(C) $15\text{ cm}$
(D) $24.5\text{ cm}$
Answer:
The correct option is (A) $7\text{ cm}$.
Step-by-Step Mathematical Evaluation:
- Let $P$ be the point of contact on the circle of centre $O$, and $Q$ be the external point.
- Tangent length: $PQ = 24\text{ cm}$.
- Distance of $Q$ from centre: $OQ = 25\text{ cm}$.
- By Theorem 10.1:
$$OP \perp PQ \implies \angle OPQ = 90^\circ$$ - By Pythagoras Theorem in right triangle $\triangle OPQ$:
$$OQ^2 = OP^2 + PQ^2$$
$$25^2 = OP^2 + 24^2$$
$$625 = OP^2 + 576$$
$$OP^2 = 625 – 576 = 49$$
$$OP = \sqrt{49} = 7\text{ cm}$$
Final Answer:
The radius of the circle is $7\text{ cm}$.
Question 2 (Page 153) [CBSE 2012, 2019 Set-2]
In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is equal to:
(A) $60^\circ$
(B) $70^\circ$
(C) $80^\circ$
(D) $90^\circ$
Answer:
The correct option is (B) $70^\circ$.
Step-by-Step Mathematical Evaluation:
- By Theorem 10.1, the tangents are perpendicular to their corresponding radii at the points of contact:
$$OP \perp TP \implies \angle OPT = 90^\circ$$
$$OQ \perp TQ \implies \angle OQT = 90^\circ$$ - In the quadrilateral $OPTQ$, the sum of all four interior angles is $360^\circ$:
$$\angle POQ + \angle OPT + \angle PTQ + \angle OQT = 360^\circ$$
$$110^\circ + 90^\circ + \angle PTQ + 90^\circ = 360^\circ$$
$$\angle PTQ + 290^\circ = 360^\circ$$
$$\angle PTQ = 360^\circ – 290^\circ = 70^\circ$$
Final Answer:
$\angle PTQ$ is $70^\circ$.
Question 3 (Page 153) [CBSE 2016, 2023 Set-3]
If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of $80^\circ$, then $\angle POA$ is equal to:
(A) $50^\circ$
(B) $60^\circ$
(C) $70^\circ$
(D) $80^\circ$
Answer:
The correct option is (A) $50^\circ$.
Step-by-Step Mathematical Evaluation:
- Determine the central angle $\angle AOB$:
In quadrilateral $OAPB$:
$$\angle OAP = 90^\circ, \quad \angle OBP = 90^\circ \quad (\text{By Theorem 10.1})$$
Since opposite angles in this tangent quadrilateral sum to $180^\circ$:
$$\angle AOB + \angle APB = 180^\circ$$
$$\angle AOB + 80^\circ = 180^\circ \implies \angle AOB = 100^\circ$$ - Join $OP$ and compare triangles $\triangle OAP$ and $\triangle OBP$:
- $OA = OB$ (Radii)
- $PA = PB$ (Theorem 10.2)
- $OP = OP$ (Common)
$$\triangle OAP \cong \triangle OBP \quad (\text{SSS Congruence})$$
- By CPCTC:
$$\angle POA = \angle POB = \frac{1}{2}\angle AOB$$
$$\angle POA = \frac{100^\circ}{2} = 50^\circ$$
Final Answer:
$\angle POA$ is $50^\circ$.
Question 4 (Page 153) [CBSE 2014, 2017, 2020 Standard]
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer: O A B P Q R S
Given Data:
$AB$ is a diameter of a circle with centre $O$. Lines $PQ$ and $RS$ are tangents drawn to the circle at the endpoints $A$ and $B$, respectively.
To Prove:
$$PQ \parallel RS$$
Proof:
- $AB$ is the diameter of the circle; therefore, $OA$ and $OB$ are radii lying along the same straight line.
- By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact:
$$OA \perp PQ \implies \angle PAB = 90^\circ \quad \text{— (Equation 1)}$$
$$OB \perp RS \implies \angle ABS = 90^\circ \quad \text{— (Equation 2)}$$ - From Equations 1 and 2:
$$\angle PAB = \angle ABS = 90^\circ$$ - These two angles form a pair of alternate interior angles for lines $PQ$ and $RS$ intersected by the transversal line $AB$.
- When alternate interior angles are equal, the lines are parallel:
$$PQ \parallel RS$$
Conclusion:
Hence Proved.
Question 5 (Page 153) [CBSE 2015, 2019 Set-1]
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer:
Given Data:
A circle with centre $O$ and a tangent line $AB$ touching the circle at point $P$. A perpendicular line $PQ$ is drawn to $AB$ at point $P$ ($PQ \perp AB$).
To Prove:
The line $PQ$ passes through the centre $O$.
Proof by Contradiction:
- Assume the contrary: suppose the perpendicular line $PQ$ does not pass through the centre $O$.
- Let another point $O’$ exist that is not coincident with $O$ such that line $O’P \perp AB$.
- By this assumption:
$$\angle O’PB = 90^\circ \quad \text{— (Equation 1)}$$ - However, $O$ is the true centre of the circle, and $P$ is the point of contact. By Theorem 10.1, the radius drawn through the point of contact is strictly perpendicular to the tangent:
$$OP \perp AB \implies \angle OPB = 90^\circ \quad \text{— (Equation 2)}$$ - Comparing Equation 1 and Equation 2:
$$\angle O’PB = \angle OPB$$ - Looking at the geometric orientation, $\angle O’PB$ is a part of $\angle OPB$. A part can equal the whole if and only if the ray $O’P$ coincides with the ray $OP$.
- Therefore, our assumption was incorrect. The perpendicular to the tangent at the point of contact must pass through the centre $O$.
Conclusion:
Hence Proved.
Question 6 (Page 153) [CBSE 2018, 2022 Term-2]
The length of a tangent from a point $A$ at distance $5\text{ cm}$ from the centre of the circle is $4\text{ cm}$. Find the radius of the circle.
Answer:
Step 1: Identify Given Data
- Distance of external point $A$ from centre $O$: $OA = 5\text{ cm}$.
- Length of tangent from point $A$ to point of contact $T$: $AT = 4\text{ cm}$.
- Let the radius of the circle be $OT = r$.
Step 2: Mathematical Formulation
By Theorem 10.1:
$$OT \perp AT \implies \angle OTA = 90^\circ$$
Step 3: Step-by-Step Substitution
Apply Pythagoras Theorem in right-angled triangle $\triangle OTA$:
$$OA^2 = OT^2 + AT^2$$
$$5^2 = r^2 + 4^2$$
$$25 = r^2 + 16$$
$$r^2 = 25 – 16 = 9$$
$$r = \sqrt{9} = 3\text{ cm}$$
Final Answer:
The radius of the circle is $3\text{ cm}$.
Question 7 (Page 153) [CBSE 2011, 2017, 2020 Standard, 2024]
Two concentric circles are of radii $5\text{ cm}$ and $3\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Step 1: Identify Given Data
- Radius of larger circle $C_1$: $R = 5\text{ cm}$.
- Radius of smaller circle $C_2$: $r = 3\text{ cm}$.
- Let $AB$ be the chord of $C_1$ that touches $C_2$ at the point of contact $P$.
Step 2: Geometric Deduction
- Because $AB$ touches the smaller circle $C_2$ at point $P$, $AB$ is tangent to $C_2$.
- By Theorem 10.1:
$$OP \perp AB \implies \angle OPA = 90^\circ$$ - By Example 1 / Standard Circle Properties: the perpendicular from the centre to a chord bisects the chord:
$$AP = PB = \frac{1}{2}AB$$
Step 3: Solve Triangle $\triangle OPA$
In right-angled triangle $\triangle OPA$:
$$OA^2 = OP^2 + AP^2$$
$$5^2 = 3^2 + AP^2$$
$$25 = 9 + AP^2$$
$$AP^2 = 16 \implies AP = 4\text{ cm}$$
Step 4: Compute Total Chord Length ($AB$)
$$AB = 2 \times AP = 2 \times 4\text{ cm} = 8\text{ cm}$$
Final Answer:
The length of the chord is $8\text{ cm}$.
Question 8 (Page 153) [CBSE 2012, 2016, 2019, 2023 Set-2]
A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that:
$$AB + CD = AD + BC$$
Answer: A B C D P Q R S
Given Data:
Quadrilateral $ABCD$ circumscribes a circle with centre $O$. The sides $AB$, $BC$, $CD$, and $DA$ touch the circle at points $P$, $Q$, $R$, and $S$, respectively.
To Prove:
$$AB + CD = AD + BC$$
Proof:
- Points $A$, $B$, $C$, and $D$ are external points from which pairs of tangents are drawn to the circle.
- By Theorem 10.2, the lengths of tangents drawn from an external point to a circle are equal:
- From point $A$:
$$AP = AS \quad \text{— (Equation 1)}$$ - From point $B$:
$$BP = BQ \quad \text{— (Equation 2)}$$ - From point $C$:
$$CR = CQ \quad \text{— (Equation 3)}$$ - From point $D$:
$$DR = DS \quad \text{— (Equation 4)}$$
- Add Equations (1), (2), (3), and (4):
$$(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)$$ - Regroup the terms on both sides to match the boundary line segments:
$$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$ - From the diagram:
$$AP + BP = AB$$
$$CR + DR = CD$$
$$AS + DS = AD$$
$$BQ + CQ = BC$$ - Substituting these sums yields:
$$AB + CD = AD + BC$$
Conclusion:
Hence Proved.
Question 9 (Page 154) [CBSE 2012, 2017, 2020 Standard, 2024 Set-3]
In Fig. 10.13, $XY$ and $X’Y’$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X’Y’$ at $B$. Prove that $\angle AOB = 90^\circ$.
Answer: O A B P Q C
Given Data:
$XY \parallel X’Y’$ are two parallel tangents touching a circle with centre $O$ at points $P$ and $Q$, meaning $POQ$ is a diameter. A third tangent $AB$ touches at point $C$, intersecting $XY$ at $A$ and $X’Y’$ at $B$.
To Prove:
$$\angle AOB = 90^\circ$$
Construction:
Join $OC$.
Step-by-Step Proof:
- Compare $\triangle OPA$ and $\triangle OCA$:
- $AP = AC$ [Tangents drawn from external point $A$, Theorem 10.2]
- $OP = OC$ [Radii of the same circle]
- $OA = OA$ [Common side]
- By SSS Congruence Criterion:
$$\triangle OPA \cong \triangle OCA$$ - By CPCTC:
$$\angle POA = \angle COA \implies \angle POC = 2\angle COA \quad \text{— (Equation 1)}$$
- Similarly, compare $\triangle OQB$ and $\triangle OCB$:
- $BQ = BC$ [Tangents drawn from external point $B$, Theorem 10.2]
- $OQ = OC$ [Radii of the same circle]
- $OB = OB$ [Common side]
- By SSS Congruence Criterion:
$$\triangle OQB \cong \triangle OCB$$ - By CPCTC:
$$\angle QOB = \angle COB \implies \angle QOC = 2\angle COB \quad \text{— (Equation 2)}$$
- Because $POQ$ is a straight line diameter:
$$\angle POC + \angle QOC = 180^\circ$$ - Substitute Equations 1 and 2 into the straight angle sum:
$$2\angle COA + 2\angle COB = 180^\circ$$
$$2(\angle COA + \angle COB) = 180^\circ$$
$$\angle COA + \angle COB = \frac{180^\circ}{2} = 90^\circ$$ - Looking at the diagram:
$$\angle COA + \angle COB = \angle AOB$$
$$\angle AOB = 90^\circ$$
Conclusion:
Hence Proved.
Question 10 (Page 154) [CBSE 2014, 2017, 2020 Standard]
Prove that 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.
Answer:
Given Data:
Tangents $PA$ and $PB$ are drawn from external point $P$ touching the circle with centre $O$ at points $A$ and $B$.
To Prove:
$$\angle APB + \angle AOB = 180^\circ$$
Step-by-Step Proof:
- By Theorem 10.1, the radius drawn through the point of contact is perpendicular to the tangent line:
$$OA \perp PA \implies \angle OAP = 90^\circ$$
$$OB \perp PB \implies \angle OBP = 90^\circ$$ - In quadrilateral $OAPB$, the sum of all four interior angles is $360^\circ$:
$$\angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ$$ - Substitute the values of the two right angles:
$$\angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ$$
$$\angle APB + \angle AOB + 180^\circ = 360^\circ$$
$$\angle APB + \angle AOB = 360^\circ – 180^\circ$$
$$\angle APB + \angle AOB = 180^\circ$$
Conclusion:
Hence proved, the angle between the two tangents and the angle subtended by the points of contact at the centre are supplementary.
Question 11 (Page 154) [CBSE 2013, 2017, 2019, 2023 Standard]
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:
Given Data:
$ABCD$ is a parallelogram circumscribing a circle with centre $O$. Its sides $AB$, $BC$, $CD$, and $DA$ touch the circle at points $P$, $Q$, $R$, and $S$, respectively.
To Prove:
$ABCD$ is a rhombus (i.e., $AB = BC = CD = DA$).
Step-by-Step Proof:
- Because $ABCD$ is a parallelogram, its opposite sides are equal:
$$AB = CD \quad \text{— (Equation 1)}$$
$$AD = BC \quad \text{— (Equation 2)}$$ - From the proof established in Question 8 (circumscribing quadrilateral identity via Theorem 10.2):
$$AB + CD = AD + BC \quad \text{— (Equation 3)}$$ - Substitute Equation 1 and Equation 2 into Equation 3:
$$AB + AB = BC + BC$$
$$2AB = 2BC$$
$$AB = BC \quad \text{— (Equation 4)}$$ - Combining Equations 1, 2, and 4:
$$AB = BC = CD = DA$$ - A parallelogram with all four sides equal is, by definition, a rhombus.
Conclusion:
Hence Proved.
Question 12 (Page 154) [CBSE 2014, 2018, 2020 Standard, 2024 Set-1]
A triangle $ABC$ is drawn to circumscribe a circle of radius $4\text{ cm}$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8\text{ cm}$ and $6\text{ cm}$ respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.
Answer: A B C D E F O
Given Data:
- Radius of incircle: $r = OD = OE = OF = 4\text{ cm}$.
- Segments of base $BC$: $BD = 8\text{ cm}$, $CD = 6\text{ cm}$.
- Total base length: $BC = BD + CD = 8 + 6 = 14\text{ cm}$.
Step 1: Express Side Lengths Using Tangent Segments
Let the circle touch $AB$ at $E$ and $AC$ at $F$. By Theorem 10.2:
- From vertex $C$: $CF = CD = 6\text{ cm}$
- From vertex $B$: $BE = BD = 8\text{ cm}$
- From vertex $A$: let $AE = AF = x\text{ cm}$
The side lengths of $\triangle ABC$ are:
$$a = BC = 14\text{ cm}$$
$$b = AC = (x + 6)\text{ cm}$$
$$c = AB = (x + 8)\text{ cm}$$
Step 2: Calculate Area of $\triangle ABC$ using Heron’s Formula
- Compute the semi-perimeter $s$:
$$s = \frac{a + b + c}{2} = \frac{14 + (x + 6) + (x + 8)}{2} = \frac{2x + 28}{2} = x + 14$$ - Compute the individual factors:
$$s – a = (x + 14) – 14 = x$$
$$s – b = (x + 14) – (x + 6) = 8$$
$$s – c = (x + 14) – (x + 8) = 6$$ - By Heron’s Formula:
$$\text{Area}(\triangle ABC) = \sqrt{s(s-a)(s-b)(s-c)}$$
$$\text{Area}(\triangle ABC) = \sqrt{(x + 14)(x)(8)(6)} = \sqrt{48x(x + 14)}\text{ cm}^2 \quad \text{— (Equation 1)}$$
Step 3: Calculate Area of $\triangle ABC$ by Summing Component Triangles
Join $OA$, $OB$, and $OC$:
$$\text{Area}(\triangle ABC) = \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB)$$
$$\text{Area} = \left(\frac{1}{2} \times BC \times r\right) + \left(\frac{1}{2} \times AC \times r\right) + \left(\frac{1}{2} \times AB \times r\right)$$
$$\text{Area} = \frac{r}{2}(BC + AC + AB) = \frac{4}{2}(2s) = 4s$$
Substitute $s = x + 14$:
$$\text{Area}(\triangle ABC) = 4(x + 14)\text{ cm}^2 \quad \text{— (Equation 2)}$$
Step 4: Equate Both Area Expressions to Solve for $x$
$$\sqrt{48x(x + 14)} = 4(x + 14)$$
Square both sides:
$$48x(x + 14) = 16(x + 14)^2$$
Since length $x + 14 \neq 0$, divide both sides by $16(x + 14)$:
$$3x = x + 14$$
$$3x – x = 14$$
$$2x = 14 \implies x = 7\text{ cm}$$
Step 5: Compute Sides $AB$ and $AC$
$$AB = x + 8 = 7 + 8 = 15\text{ cm}$$
$$AC = x + 6 = 7 + 6 = 13\text{ cm}$$
Final Answer:
The lengths of the sides are $AB = 15\text{ cm}$ and $AC = 13\text{ cm}$.
Question 13 (Page 154) [CBSE 2015, 2018, 2020 Standard, 2023]
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer: A B C D O
Given Data:
A quadrilateral $ABCD$ circumscribes a circle with centre $O$. The circle touches the sides $AB$, $BC$, $CD$, and $DA$ at points $P$, $Q$, $R$, and $S$, respectively.
To Prove:
$$\angle AOB + \angle COD = 180^\circ$$
$$\angle BOC + \angle AOD = 180^\circ$$
Construction:
Join the centre $O$ to the vertices $A, B, C, D$ and to the points of contact $P, Q, R, S$.
Step-by-Step Proof:
- Consider the two right-angled triangles $\triangle OAP$ and $\triangle OAS$:
- $AP = AS$ [Theorem 10.2]
- $OP = OS$ [Radii of the same circle]
- $OA = OA$ [Common hypotenuse]
- $\triangle OAP \cong \triangle OAS$ [By SSS or RHS Congruence]
- By CPCTC:
$$\angle 1 = \angle 8$$ - Similarly, applying the same congruency to the other pairs:
$$\angle 2 = \angle 3$$
$$\angle 4 = \angle 5$$
$$\angle 6 = \angle 7$$ - The sum of all angles formed around the central point $O$ is $360^\circ$:
$$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$$ - Group the angles corresponding to opposite sides $AB$ and $CD$:
$$(\angle 1 + \angle 8) + (\angle 2 + \angle 3) + (\angle 4 + \angle 5) + (\angle 6 + \angle 7) = 360^\circ$$
$$2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^\circ$$
$$2(\angle 1 + \angle 2) + 2(\angle 5 + \angle 6) = 360^\circ$$
$$\angle 1 + \angle 2 + \angle 5 + \angle 6 = 180^\circ$$ - From the diagram, notice that $\angle 1 + \angle 2 = \angle AOB$ and $\angle 5 + \angle 6 = \angle COD$:
$$\angle AOB + \angle COD = 180^\circ$$ - Because the sum of all four component angles is $360^\circ$:
$$\angle BOC + \angle AOD = 360^\circ – (\angle AOB + \angle COD) = 360^\circ – 180^\circ = 180^\circ$$
Conclusion:
Hence Proved.
Master High-Yield Board FAQs (Rank Math FAQ Schema Ready)
What is the geometric difference between a secant and a tangent?
A secant is an extended straight line that cuts through a circle at two distinct points, creating an internal chord. In contrast, a tangent touches the circle at only one single point, known as the point of contact, and remains entirely outside the circle without crossing into its interior.
How many tangents can be drawn to a circle from a point inside it?
Zero tangents can be drawn from a point located inside a circle. Any straight line passing through an interior point intersects the circle’s circumference at two distinct points, making every such line a secant.
How many tangents can be drawn from a point lying on the circle?
Exactly one unique tangent can be drawn to a circle from a point lying directly on its circumference. By Theorem 10.1, this tangent line is strictly perpendicular to the radius connecting the centre to that point of contact.
How many tangents can be drawn from a point outside the circle?
From any point situated outside a circle, exactly two distinct tangents can be drawn. By Theorem 10.2, these two tangents have equal lengths from the external point to their respective points of contact.
Why is the radius perpendicular to the tangent at the point of contact?
The point of contact is the closest point on the tangent line to the circle’s centre. Every other point on the tangent lies outside the circle, meaning its distance from the centre is greater than the circle’s radius. Because the shortest line segment connecting a point to a straight line is always perpendicular, the radius drawn to the point of contact must be perpendicular to the tangent.
Can two tangents drawn to a circle be parallel to each other?
Yes, two tangents can be parallel, but only if they are drawn at the opposite endpoints of a diameter. Because the diameter forms a straight transversal line perpendicular to both tangents at their points of contact, the resulting alternate interior angles each equal $90^\circ$, making the tangents parallel.
What is the circumscribed quadrilateral property for a circle?
When a quadrilateral circumscribes a circle, the sum of the lengths of one pair of opposite sides equals the sum of the lengths of the other pair ($AB + CD = AD + BC$). This follows directly from Theorem 10.2, as each vertex acts as an external point with equal pairs of tangent segments.
Why does a parallelogram circumscribing a circle have to be a rhombus?
A parallelogram circumscribing a circle satisfies $AB + CD = AD + BC$. Because opposite sides of a parallelogram are equal ($AB = CD$ and $AD = BC$), the equation simplifies to $2AB = 2BC$, which means $AB = BC$. A parallelogram with adjacent sides equal is, by definition, a rhombus.
What criteria are used to prove Theorem 10.2 in CBSE board exams?
Theorem 10.2 is proved using the RHS (Right Angle-Hypotenuse-Side) congruence criterion. In triangles $\triangle OQP$ and $\triangle ORP$, $\angle OQP = \angle ORP = 90^\circ$ (by Theorem 10.1), $OP = OP$ (common hypotenuse), and $OQ = OR$ (radii of the same circle). SSS congruence can also be used if the lengths are shown equal using the Pythagorean identity.
How do opposite sides of a circumscribed quadrilateral subtend angles at the centre?
The opposite sides of a circumscribed quadrilateral always subtend supplementary angles at the circle’s centre ($\angle AOB + \angle COD = 180^\circ$). This is proved by connecting the centre to all four vertices and all four points of contact, creating eight pairwise congruent triangles whose central angles sum to $360^\circ$.
What happens to the angle between two tangents as the external point moves farther away?
As an external point moves farther from the circle’s centre, the angle between the two tangents decreases towards $0^\circ$. At the same time, the central angle between the radii drawn to the points of contact increases towards $180^\circ$, because these two angles remain supplementary ($\angle APB + \angle AOB = 180^\circ$).
How do you find the area of a circumscribed triangle using its inradius?
The area of a triangle circumscribed around a circle equals the product of its semi-perimeter and the inradius ($\text{Area} = r \cdot s$). This formula is derived by dividing the triangle into three component triangles ($\triangle OBC$, $\triangle OCA$, $\triangle OAB$) sharing the circle’s centre $O$ as an apex and having inradius $r$ as their common height.
Must students draw diagrams for geometric proofs in Class 10 board exams?
Yes. CBSE official marking schemes mandate a neat, labelled geometric diagram for all circle questions. An answer presented without a corresponding diagram can be penalized up to $1$ mark, and subsequent steps in the proof may be considered unsupported if labels are not clearly defined visually.
What is a cyclic quadrilateral and how does it relate to tangents?
A cyclic quadrilateral is a quadrilateral whose four vertices all lie on a single circle. In tangent problems, when two tangents are drawn from an external point $P$ touching at $A$ and $B$, the quadrilateral $OAPB$ is cyclic because its opposite angles at $A$ and $B$ are both right angles, summing to $180^\circ$.
What common mistake should students avoid in Exercise 10.2 Question 12?
The most common mistake in Question 12 is incorrectly computing the semi-perimeter or mishandling the algebraic expansion when equating Heron’s formula to $r \cdot s$. Students often forget to divide the perimeter by 2, leading to incorrect values for $s$ and invalid quadratic terms. Always divide the perimeter by 2 first: $s = x + 14$.