NCERT Solutions for Class 10 Maths Chapter 10: Circles — Free PDF

Chapter 10 focuses on tangents to a circle — their properties and applications. This is a proof-heavy chapter that builds directly on the circle theorems you learnt in Class 9 (Chapter 9) and the triangle congruence skills from Class 9 (Chapter 7).

Unlike Class 9 circles (which focused on chords, arcs, and cyclic quadrilaterals), Class 10 circles is entirely about tangent lines — lines that touch a circle at exactly one point. The two key theorems in this chapter are used in almost every problem.

The chapter has 2 exercises covering:
- Exercise 10.1: Conceptual questions about the number of tangents from a point to a circle (fill-in-the-blank type, 1-mark questions).
- Exercise 10.2: Properties of tangents, proofs, and numerical problems involving Pythagoras theorem (3-5 mark questions).

This chapter typically carries 4-6 marks in the CBSE board exam, often as one proof-based question (3 marks) and one numerical problem (2-3 marks). The proofs are highly predictable — mastering the two main theorems and their applications ensures full marks.

Tangent: A line that touches a circle at exactly one point. This single point is called the point of contact (or point of tangency).

Secant: A line that intersects a circle at two points. A secant passes through the interior of the circle.

Number of Tangents from a Point:
- If the point is inside the circle: zero tangents (every line through the point intersects the circle at two points).
- If the point is on the circle: exactly one tangent (perpendicular to the radius at that point).
- If the point is outside the circle: exactly two tangents (and they are equal in length).

Point of Contact: The single point where a tangent line touches the circle.

Tangent Segment: The part of the tangent line between the external point and the point of contact.

Chord of Contact: If two tangents are drawn from an external point $P$ to a circle, touching it at $A$ and $B$, then $AB$ is called the chord of contact.

Theorem 10.1: The tangent-radius perpendicularity theorem.

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

This means $\angle OAP = 90^\circ$. This right angle is the foundation for applying the Pythagoras theorem in tangent problems.

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Theorem 10.2: The equal tangent length theorem.

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

Proof (frequently asked in exams):

In $\triangle OPA$ and $\triangle OPB$:
- $OA = OB$ (radii of the same circle)
- $OP = OP$ (common side)
- $\angle OAP = \angle OBP = 90^\circ$ (Theorem 10.1)

By RHS congruence: $\triangle OPA \cong \triangle OPB$

By CPCT: $PA = PB$ $\blacksquare$

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Useful Corollaries (derived from Theorem 10.2):

1. $OP$ bisects $\angle APB$ (the angle between the tangents): $\angle OPA = \angle OPB$.
2. $OP$ bisects the chord of contact $AB$ at right angles.
3. $\angle OPA = \angle OPB$ and $\angle AOB + \angle APB = 180^\circ$.
4. If a circle is inscribed in a triangle with sides $a, b, c$ and semi-perimeter $s = \dfrac{a+b+c}{2}$, the tangent lengths from each vertex are: $s - a$, $s - b$, $s - c$.

These problems use the tangent-radius perpendicularity and equal tangent length properties.

Additional problems matching the difficulty of CBSE board exams.

**Example 1: A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.**

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

By equal tangent lengths:
- From $A$: $AP = AS$
- From $B$: $BP = BQ$
- From $C$: $CQ = CR$
- From $D$: $DR = DS$

$AB + CD = (AP + PB) + (CR + RD) = AS + BQ + CQ + DS$

$AD + BC = (AS + SD) + (BQ + QC) = AS + DS + BQ + CQ$

Both expressions are equal:

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**Example 2: If tangents $PA$ and $PB$ from an external point $P$ make an angle of $80^\circ$, find $\angle AOB$.**

Solution:
Using $\angle AOB + \angle APB = 180^\circ$:

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**Example 3: From a point $P$, two tangents $PA$ and $PB$ are drawn to a circle with centre $O$ and radius $r$. If $OP = 2r$, find $\angle APB$.**

Solution:
In right $\triangle OAP$: $\sin(\angle OPA) = \dfrac{OA}{OP} = \dfrac{r}{2r} = \dfrac{1}{2}$.

$\angle OPA = 30^\circ$.

Since $OP$ bisects $\angle APB$: $\angle APB = 2 \times 30^\circ = 60^\circ$.

Mistake 1: Forgetting to draw the radius and mark the right angle.
The right angle between the tangent and radius is the starting point for almost every problem. Always draw the radius to the point of contact and mark $90^\circ$.

Mistake 2: Confusing tangent length with chord length.
The tangent length ($PA$) is the distance from the external point to the point of contact. The chord of contact ($AB$) is different. Make sure you know which is being asked.

Mistake 3: Not using the tangent length system for inscribed circle problems.
When a circle is inscribed in a triangle, the tangent lengths from each vertex follow the pattern: $s - a$, $s - b$, $s - c$ (where $s$ is the semi-perimeter). This saves significant computation.

**Mistake 4: Forgetting that $\angle AOB + \angle APB = 180^\circ$.**
This supplementary angle relationship is frequently tested and easy to forget. Always check if it applies in the problem.

Mistake 5: Not stating Theorem 10.1 or 10.2 when using them in proofs.
In CBSE exams, you must cite the theorem by name or statement when using it. Simply writing "tangent is perpendicular" without referencing Theorem 10.1 may cost marks.

Q1. Two tangents from an external point $P$ touch a circle of radius $5$ cm. If the distance from $P$ to the centre is $13$ cm, find the tangent length.

Answer: $PA = \sqrt{OP^2 - r^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

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Q2. If tangents $PA$ and $PB$ from $P$ to a circle with centre $O$ make $\angle APB = 60^\circ$, find $\angle AOB$.

Answer: $\angle AOB = 180^\circ - 60^\circ = 120^\circ$.

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Q3. A circle is inscribed in $\triangle ABC$ with $AB = 14$ cm, $BC = 13$ cm, $CA = 15$ cm. Find the tangent lengths from each vertex.

Answer: $s = (14 + 13 + 15)/2 = 21$. Tangent from $A = s - a = 21 - 13 = 8$ cm. From $B = s - b = 21 - 15 = 6$ cm. From $C = s - c = 21 - 14 = 7$ cm.

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Q4. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Answer: Assume the tangent is not perpendicular. Then there exists a point $Q$ on the tangent closer to $O$ than the point of contact $P$. But $Q$ must be outside the circle (since the tangent touches the circle at only $P$), so $OQ > OP = r$. This contradicts $OQ < OP$. Therefore the tangent must be perpendicular to the radius. $\blacksquare$

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Q5. Two concentric circles have radii $13$ cm and $5$ cm. Find the length of the chord of the outer circle that touches the inner circle.

Answer: Let the chord touch the inner circle at $M$. $OM = 5$ cm (inner radius). $OA = 13$ cm (outer radius). $AM = \sqrt{169 - 25} = 12$ cm. Chord $= 2 \times 12 = 24$ cm.

• This chapter revolves around two key theorems: tangent $\perp$ radius (Theorem 10.1) and equal tangent lengths from an external point (Theorem 10.2).
  - The proof of Theorem 10.2 using RHS congruence is a very commonly asked board exam question.
  - $\angle AOB + \angle APB = 180^\circ$ (tangent angle + central angle = supplementary).
  - For a circle inscribed in a triangle: tangent from vertex $= s -$ (opposite side), where $s$ is the semi-perimeter.
  - For a quadrilateral circumscribing a circle: $AB + CD = AD + BC$.
  - Pythagoras theorem is your main computational tool (applied in the right triangle formed by radius, tangent, and line to centre).
  - Always draw the radius to the point of contact and mark the $90^\circ$ angle as the first step.
  - This chapter connects directly to Class 9 circles (chord/arc properties) and to Chapter 11 (Areas Related to Circles).

Practise circle problems on SparkEd for detailed solutions to tangent-based problems!