Class 10 Maths Chapter 10: Circles Notes & Formulas

Exam Weightage & Blueprint

Total: ~6-8 Marks

This chapter falls under Unit IV: Geometry (15 marks total). As per the latest syllabus, focus is on: Tangent to a circle at point of contact. Proof that tangent is perpendicular to radius, and proof that tangents from an external point are equal.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Tangent-Radius angle ($90^^\circ$), Lengths of tangents
Short Answer	2 or 3	Very High	Proof of Theorem 10.2, Concentric circles, Finding lengths (Pythagoras)
Long Answer	4 or 5	Medium	Quadrilaterals circumscribing circles, Complex proofs

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Secant vs Tangent Definition: A tangent is the limiting position of a secant when the two endpoints of its corresponding chord coincide. There can be at most 2 parallel tangents to a circle for a given secant.
Diameter Tangents: Tangents drawn at the endpoints of a diameter are always parallel.
Supplementary Angles Rule: The angle between two tangents drawn from an external point and the angle subtended by the line segment joining the points of contact at the centre are supplementary: $$\angle PTQ + \angle POQ = 180^\circ$$
Circumscribed Quadrilateral Properties:
• $AB + CD = AD + BC$ (Sum of opposite sides are equal).
• Opposite sides subtend supplementary angles at the centre: $\angle AOB + \angle COD = 180^\circ$.
• A parallelogram circumscribing a circle is a Rhombus.
Incircle Radius of Right Triangle: $r = \frac{a + b - c}{2} = \frac{AB + BC - AC}{2}$.

⏰ Last 24-Hour Checklist

Theorem 10.1: Radius $\perp$ Tangent.
Theorem 10.2: Lengths of tangents from external point are equal.
No Tangent: From a point inside the circle.

One Tangent: At a point on the circle.
Two Tangents: From a point outside the circle.
Quad Property: $AB+CD = AD+BC$ (for circumscribing quad).

📐 Important Theorems

Theorem 10.1: Tangent-Radius Perpendicularity

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

Figure 10.1: Tangent XY is Perpendicular to Radius OP at Point of Contact

Implication: In any problem with a tangent, look for the triangle formed by the radius, tangent, and the line joining the center to the external point. It will be a Right-Angled Triangle. Use Pythagoras Theorem: $$ (Hypotenuse)^2 = (Radius)^2 + (Tangent)^2 $$

Theorem 10.2: Tangents from External Point

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

Figure 10.2: Tangents PQ and PR Drawn From External Point P are Equal

Proof Outline (Often asked for 3 Marks):

1. Join $OP, OQ, OR$.

2. In $\Delta OQP$ and $\Delta ORP$:

- $\angle OQP = \angle ORP = 90^^\circ$ (Radius $\perp$ Tangent)

- $OQ = OR$ (Radii)

- $OP = OP$ (Common)

3. $\therefore \Delta OQP \cong \Delta ORP$ (RHS Rule).

4. $\Rightarrow PQ = PR$ (CPCT).

Key Deductions from Theorem 10.2

1. Length of Tangent

If $d$ is the distance to the external point and $r$ is the radius, length $l$ is:

$$ l = \sqrt{d^2 - r^2} $$

2. Circumscribing Quadrilateral

If a quadrilateral ABCD circumscribes a circle, sum of opposite sides is equal:

$$ AB + CD = AD + BC $$

Concept: Number of Tangents

Point Position	Number of Tangents	Diagram Representation
Inside Circle	0	Secant (intersects at 2 points)
On the Circle	1	Unique Tangent
Outside Circle	2	Two tangents of equal length

CBSE 2026–27 Syllabus Status: Circles

Unit IV: Geometry • 6–8 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

Solved Examples (Board Marking Scheme)

Q1. Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact. (3 Marks)

Step 1: Diagram & Given (0.5 Mark)

Let $C_1$ be the larger circle, $C_2$ the smaller circle. Chord $AB$ of $C_1$ touches $C_2$ at $P$. Center $O$.

Step 2: Tangent Property (1 Mark)

Since $AB$ is a tangent to $C_2$ at $P$ and $OP$ is the radius, $OP \perp AB$ (Theorem 10.1).

Step 3: Chord Property (1.5 Marks)

For the larger circle $C_1$, $AB$ is a chord and $OP \perp AB$.

We know that the perpendicular from the center to a chord bisects the chord.

$\therefore AP = BP$. Hence Proved.

Q2. Two tangents TP and TQ are drawn to a circle with center O from an external point T. Prove that $\angle PTQ = 2\angle OPQ$. (3 Marks)

Step 1: Isosceles Triangle Property (1 Mark)

Let $\angle PTQ = \theta$.

Since $TP = TQ$ (Theorem 10.2), $\Delta TPQ$ is isosceles.

$\therefore \angle TPQ = \angle TQP = \frac{1}{2}(180^° - θ) = 90^° - \frac{θ}{2}$.

Step 2: Tangent Property (1 Mark)

We know $\angle OPT = 90^^\circ$ (Radius $\perp$ Tangent).

Step 3: Final Calculation (1 Mark)

$\angle OPQ = \angle OPT - \angle TPQ = 90^° - (90^° - \frac{θ}{2}) = \frac{θ}{2}$.

$\therefore \angle OPQ = \frac{1}{2}\angle PTQ \Rightarrow \angle PTQ = 2\angle OPQ$.

Q3. PQ is a chord of length 8 cm of a circle of radius 5 cm. Tangents at P and Q intersect at T. Find length TP. (4 Marks)

Step 1: Geometry Setup (1 Mark)

Join $OT$ intersecting $PQ$ at $R$. $\Delta TPQ$ is isosceles, so $OT \perp PQ$ and bisects it.

$PR = RQ = 4$ cm.

Step 2: Find OR (1 Mark)

In $\Delta PRO$, $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3$ cm.

Step 3: Similarity/Pythagoras (2 Marks)

Let $TP = x$. In $\Delta PRT$, $x^2 = TR^2 + 4^2$. In $\Delta OPT$, $OT^2 = x^2 + 5^2$.

Alternatively, $\Delta PRO \sim \Delta TRP$ (AA Similarity).

$\frac{TP}{PO} = \frac{RP}{RO} \Rightarrow \frac{x}{5} = \frac{4}{3} \Rightarrow x = \frac{20}{3}$ cm.

Previous Year Questions (PYQs)

2023 (1 Mark): From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. Find the radius.
Ans: Radius $\perp$ Tangent. Use Pythagoras: $r = $\sqrt{25^2 - 24^2}$ = $\sqrt{49}$ = 7$ cm.

2020 (4 Marks): Prove that the parallelogram circumscribing a circle is a rhombus.
Hint: Use property $AB+CD=AD+BC$. For parallelogram, $AB=CD$ and $AD=BC$. Thus $2AB=2AD \Rightarrow AB=AD$. Adjacent sides equal \implies Rhombus.

2019 (4 Marks): 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 center.
Ans: In Quad OAPT, $\angle P = \angle T = 90^^\circ$. Sum of angles = $360^^\circ$. $\Rightarrow \angle AOB + \angle APB = 180^^\circ$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Identifying Hypotenuse: In $\Delta OPT$ (where P is point of contact), the hypotenuse is OT (Center to External Point), NOT the tangent or radius.

Tangents Length: Don't forget that tangents are equal ONLY from an external point.

Diagrams: Losing marks for not drawing or labelling diagrams in proof questions.

💡 Scoring Tips

Mark $90^^\circ$: Always mark the radius-tangent angle as $90^^\circ$ on your diagram first.

Circumscribing Quads: Remember the property: Sum of opposite sides is equal ($AB+CD = AD+BC$).

Step-wise Proofs: Write "Given", "To Prove", "Construction", and "Proof" clearly for 3-4 mark questions.

📝 More Solved Board Questions

Q4. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. 3 Marks

Sol. Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

We know that tangents from an external point are equal.

$AP = AS$ ...(1), $BP = BQ$ ...(2), $CR = CQ$ ...(3), $DR = DS$ ...(4)

Adding (1), (2), (3), and (4):

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

$AB + CD = AD + BC$. Hence Proved.

🎯 Board Pattern (2018–2025): Questions involving "quadrilateral circumscribing a circle" or "parallelogram as a rhombus" are extremely common. Always start by writing the four tangent equality equations.

📋 Board Revision Checklist

✅ Tangent at any point is $\perp$ to radius (Theorem 10.1)
✅ Tangents from an external point are equal (Theorem 10.2)
✅ Tangents from external point subtend equal angles at the center
✅ They are equally inclined to the line joining center to the point
✅ Number of tangents: Inside (0), On (1), Outside (2)
✅ In concentric circles, chord of larger circle is bisected by smaller radius
✅ If $AB+CD = AD+BC$, then quad ABCD circumscribes a circle

💡 Exam Tip:
When proving Theorem 10.2, remember to use RHS congruence criteria. Mentioning the specific congruence rule is often worth 0.5 marks.

CBSE Section E • 4 Marks

Case Study: Circular Traffic Roundabout Design

Scenario: City transport planners are laying out a circular traffic island of radius 5 m. A traffic surveillance camera post is erected at point P located 13 m away from the center O of the roundabout. Two straight security cordons PA and PB are laid tangentially from camera point P touching the boundary of the circular island at points A and B.

Q1: What is the measure of angle $\angle OAP$ between the radius OA and the tangent cordon PA? 1 Mark

Q2: Calculate the length of the tangent cordon PA. 1 Mark

Q3: Find the perimeter of quadrilateral PAOB. (OR) Find the area of quadrilateral PAOB. 2 Marks

📝 10-Mark Self-Assessment Test

Time: 25 Mins

Solve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.

Q1 (1 Mark - MCQ): From an external point $P$, tangents $PA$ and $PB$ are drawn to a circle with centre $O$. If $\angle APB = 70^^\circ$, then $\angle AOB$ is equal to:
(a) $110^^\circ$ (b) $70^^\circ$ (c) $140^^\circ$ (d) $90^^\circ$

Q2 (2 Marks - VSA): Two concentric circles are of radii $5 cm$ and $3 cm$. Find the length of the chord of the larger circle which touches the smaller circle.

Q3 (3 Marks - SA): Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Q4 (4 Marks - LA): 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. Furthermore, if the tangents are inclined at an angle of $60^^\circ$ and radius is $3 cm$, find the length of each tangent.

Sol 1: In quadrilateral $OAPB$, $\angle OAP = \angle OBP = 90^^\circ$.
$$\angle AOB + \angle APB = 180^^\circ \implies \angle AOB = 180^^\circ - 70^^\circ = 110^^\circ$$ Correct Option: (a) $110^^\circ$ [1 Mark]

Sol 2: Let chord $AB$ of larger circle touch smaller circle at $P$. Center is $O$.
$OP \perp AB$ (radius $\perp$ tangent), $OP = 3 cm$, $OA = 5 cm$.
In right $\triangle OPA$: $AP = \sqrt{OA^2 - OP^2} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4 cm$ [1 Mark].
Perpendicular from center bisects the chord $\implies AB = 2 \times AP = 2 \times 4 = 8 cm$ [1 Mark].
Answer: Length of chord $= 8 cm$ [2 Marks]

Sol 3: Let $AB$ be a diameter of a circle with centre $O$. Let lines $PQ$ and $RS$ be tangents at $A$ and $B$ respectively.
- Since $OA \perp PQ$ at point of contact $A \implies \angle OAP = 90^^\circ$ (or $\angle BAP = 90^^\circ$) [1 Mark].
- Since $OB \perp RS$ at point of contact $B \implies \angle OBS = 90^^\circ$ (or $\angle ABS = 90^^\circ$) [1 Mark].
Now, $\angle BAP = \angle ABS = 90^^\circ$. These are alternate interior angles.
Since alternate interior angles are equal, $PQ \parallel RS$. Hence Proved. [3 Marks]

Sol 4:
Part 1 (Proof): In quadrilateral $OAPB$ formed by center $O$, points of contact $A, B$, and external point $P$:
$\angle OAP = 90^^\circ$ and $\angle OBP = 90^^\circ$ (Radius $\perp$ Tangent).
Sum of angles of quadrilateral $= 360^^\circ \implies \angle APB + \angle AOB + 90^^\circ + 90^^\circ = 360^^\circ \implies \angle APB + \angle AOB = 180^^\circ$. Thus they are supplementary [2 Marks].
Part 2 (Calculation): Line $OP$ bisects $\angle APB \implies \angle APO = \frac{60^^\circ}{2} = 30^^\circ$.
In right $\triangle OAP$: $\tan 30^^\circ = \frac{OA}{AP} \implies \frac{1}{\sqrt{3}} = \frac{3}{AP} \implies AP = 3\sqrt{3} cm$ [2 Marks].
Answer: Tangent length $= 3\sqrt{3} cm$ [4 Marks]

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. How many tangents can a circle have at most?

2. The distance between two parallel tangents of a circle of radius 4 cm is:

3. If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of $80^^\circ$, then $\angle POA$ is:

4. A line intersecting a circle in two points is called a:

5. If the angle between two radii of a circle is $110^^\circ$, then the angle between the tangents at the ends of the radii is:

Frequently Asked Questions: Class 10 Maths Chapter 10

What is the marks weightage of Circles in Class 10 Maths?

Circles carries 4 to 6 marks in Unit 4: Geometry, usually including 1 MCQ on tangent properties and a 3-mark or 5-mark theorem proof or rider question.

Which theorems in Chapter 10 have compulsory proofs?

Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. Both proofs are high-yield board exam questions.

How many tangents can be drawn to a circle from a point?

From a point inside the circle: 0 tangents. From a point on the circle: exactly 1 tangent. From a point outside the circle: exactly 2 tangents of equal length.

What is the cyclic quadrilateral property related to tangents?

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 (they add up to 180 degrees).

Chapter 10 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 10: Circles

Overview