Circles Class 10
Master Circles Class 10 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
Filter
NCERT Solutions
Circles Class 10 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 10.1
4 questionsEx 10.1, 1
teachoo.com
Ex 10.1,1
How many tangents can a circle have?
Tangent is a line that intersects the circle at one point
There are infinite number of points on circle
At every point, there is one tangent
Hence, there are infinite number of tangents in a circle
Ex 10.1, 2
teachoo.com
Ex 10.1,2
Fill in the blanks
(i) A tangent to a circle intersects it in point (s).
One point
. .
Xx P Y
B
Note only there can be one tangent at point P i.e. tangent XY
If we try to make more than one line at point P example AB,
it becomes a secant (as it intersects at more than one point)
Ex 10.1, 3 (MCQ)
Ex 10.1,3 teachoo.com
A tangent PQ at a point P of a circle of radius 5 cm meets a line
through the centre O at a point Q so that OQ = 12 cm. Length PQ is :
(A} 12 cm (B) 13 cm (C} 8.5m (D) ¥119 cm.
Given OP = radius = 5 cm
& 0Q=12cm
> 12cm
Pp Q
Since PQ is a tangent,
OP L PQ | (Tangent at any point of circle is perpendicular
to the radius through point of contact}
So, Z OPQ = 90°
Hence, AOPQ is a right triangle
Ex 10.1, 4
teachoo.com
Ex 10.1,4
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.
Let the given line be AB A B
And circle be with centre O
Cc Q D
E R F
Note AB II CD II EF
Here CD is a secant
(intersecting circle at 2 points P and Q)
And
EF is a tangent
(intersecting circle at R)
Ex 10.2
13 questionsEx 10.2, 1 (MCQ)
Ex 10.2,1 teachoo.com
Choose the correct option and give justification.
From a point Q, the length of the tangent to a circle is 24 cm and
the distance of Q from the centre is 25 cm. The radius of the circle is
(A}7 cm (B) 12 cm (C} 15cm (D) 24.5 cm
Tangent = XY
Point of contact = B cm
Length of the tangent to a circle = 24 cm Se
ie. PQ= 24cm X p 24cm Q Y
&0Q=25cm
To find: Radius of circle i.e. OP
Solution:
Since XY is tangent,
to the radius through point of contact)
Ex 10.2, 2 (MCQ)
Ex 10.2,2 teachoo.com
Choose the correct option and give justification.
In figure, if TP and TQ are the two tangents to a circle with centre O
so that ZPOQ = 110°, then 2 PTQ is equal to
(A) 60° (B) 70° (C} 80° (D) 90° T
Given 2 POQ= 110° ae)
Ly de
0)
Here TP is a tangent.
So, OP L TP | (Tangent at any point of circle is
perpendicular to the radius through point of contact}
Hence, Z OPT = 90°
Similarly , TQ is a tangent.
So, OQ 1 TQ | (Tangent at any point of circle is perpendicular
to the radius through point of contact)
Hence, Z OQT = 90°
Ex 10.2, 3 (MCQ)
Ex 10.2,3 teachoo.com
Choose the correct option and give justification.
If tangents PA and PB from a point P to a circle with centre O are
inclined to each other at angle of 80°, then ZPOA is equal to
(A) 50° (B) 60° (Cc) 70° (D) 80°
Given: PA and PB are tangents to circle A
CA
& Z APB = 80° T \
Pear
To find: 2 POA SC /
Construction: Join OA,OB & OP
Proof:
Since PA is tangent,
OA L PA {Tangent at any point of circle is perpendicular
to the radius through point of contact)
-. Z OAP = 90°
Ex 10.2, 4
teachoo.com
Ex 10.2,4
Prove that the tangents drawn at the ends of a diameter of a circle
are parallel. A
P Q
Given: A circle with center O
And diameter AB
Let PQ be the tangent at point A R B 5
& RS be the tangent at point B
To prove: PQ || RS
Proof:
Since PQ is a tangent at point A
OA LPQ (Tangent at any point of circle is perpendicular
to the radius through point of contact)
Z OAP = 90° (1)
Ex 10.2, 5
Ex 10.2,5 teachoo.com
Prove that the perpendicular at the point of contact to the tangent
to a circle passes through the centre.
Given: Let us assume a circle with centre O
& AB be the tangent intersecting circle at point P
To prove: OP 1 AB
Proof:
A P B
We know that
Tangent of circle is perpendicular to radius at point of contact
Hence, OP | AB | (Tangent at any point of circle is perpendicular
to the radius through point of contact)
So, Z OPB = 90° (1)
Now lets assume some point X ,
such that XP L AB
Hence, Z XPB = 90° (2)
Ex 10.2, 6
Ex 10.2,6 teachoo.com
The length of a tangent from a point A at distance 5 cm from the
centre of the circle is 4 cm. Find the radius of the circle.
B
'
. . . 4cm
Given: Let the circle be with centre O -T \
AB is the tangent from point A A 5
c
Length of tangent = AB = 4cm
Also, distance of point from circle = 5 cm
Hence OA = 5cm
To find: Radius i.e. OB
Solution:
Since AB is tangent
Hence OB 1 AB (Tangent at any point of circle is perpendicular
to the radius through point of contact)
- Z OBA = 90°
Ex 10.2, 7
Ex 10.2,7 teachoo.com
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.
Given: Let two concentric circles be C, & C, with center O
AB be chord of the larger circle C, Cy
which touches the smaller circle C, at point P
To find: Length of AB bs
ft”
ANP O/B
Solution:
Connecting OP, OA and OB
OP = Radius of smaller circle = 3 cm
OA = OB = Radius of larger circle = 5 cm
Since AB is tangent to circle C,
OP | AB (Tangent at any point of circle is perpendicular
to the radius through point of contact)
- Z OPA = Z OPB = 90°
Ex 10.2, 8
Ex 10.2,8 teachoo.com
A quadrilateral ABCD is drawn to circumscribe a circle (see figure).
Prove that AB + CD = AD + BC R Cc
D.
Given : Let ABCD be the quadrilateral
circumscribing the circle with centre O. Q
Ss
The quadrilateral touches the circle
at points P.Q,R and S A? B
To prove: AB+ CD =AD + BC
Proof:
From theorem 10.2, lengths of tangents drawn from external point
are equal
Hence, AP = AS .(1)
BP =BQ (2)
CR=CQ ...(3)
DR=DS (4)
Ex 10.2, 9
Ex 10.2,9 teachoo.com
In figure, 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
Aand X’Y’at B. Prove that ZAOB = 90°.
Given : XY is a tangent at point P x P 7 y
and X’Y’ is a tangent at point Q YY
And XY || X’Y’ OF c
AB is a tangent at point C x’ 5 y’
To prove: 2 AOB = 90°
Proof: Join OC
For tangent AB & Radius OC
OC | AB (Tangent at any point of circle is perpendicular
to the radius through point of contact)
So, Z ACO = 2 BCO = 90°
Ex 10.2, 10
Ex 10.2,10 teachoo.com
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.
cA
Given:
A circle with center O. Px) ()
Tangents PA and PB drawn from external point P
*,
To prove: 2 APB + Z AOB = 180°
Proof:
Since PA is tangent, Since PB is tangent,
OA 1 PA OB 1 PB
(Tangent at any point of circle (Tangent at any point of circle
is perpendicular to the radius is perpendicular to the radius
through point of contact} through point of contact)
: Z OAP = 90° - Z OBP = 90°
Ex 10.2, 11
teachoo.com
Ex 10.2,11
Prove that the parallelogram circumscribing a circle is a rhombus.
D R c
A circle with centre O. s
A parallelogram ABCD touching the a
circle at points P.Q,R and § A
P
To prove: ABCD is a rhombus
Proof:
A rhombus is a parallelogram with all sides equal,
So, we have to prove all sides equal
In parallelogram ABCD,
AB=CD&AD=BC = (Opposite sides of parallelogram are equal) ...(1)
Ex 10.2, 12
Ex 10.2,12 teachoo.com
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such
that the segments BD and DC into which BC is divided by the point of
contact D are of lengths 8 cm and 6 cm respectively (see figure). Find
the sides AB and AC.
A
Given: A circle with centre O
with OD = radius = 4.cm ,
Let A ABC circumscribe the circle E \ F
Oo
Also, BD = 8 cm abr
&CD=6cm UN 4cm
Cotai—Sp<—— fem —— 8
To find: AB & AC
Construction: Join OA, OC& OB
Let AC, AB intersect circle at E & F respectively
Solution:
Ex 10.2, 13
Ex 10.2,13 teachoo.com
Prove that opposite sides of a quadrilateral circumscribing a circle
subtend supplementary angles at the centre of the circle.
R
D, C
Nae N77
Given : Let ABCD be the quadrilateral
circumscribing the circle with centre O. S
@
ABCD touches the circle at points P.Q,R and S MLN Q
To prove: P
Opposite sides subtend supplementary angles at centre
i.e. 2 AOB + Z COD = 180°
& Z AOD + Z BOC = 180°
Construction: Join OP, OQ, OR & OS
Proof:
Examples
3 questionsExample 1
Example 1(Method 1) teachoo.com
Prove that in two concentric circles, the chord of the larger circle
which touches the smaller circle is bisected at the point of contact.
Given: Let two concentric circles be C, & C,
with center O Cc,
AB be chord of the larger circle C, A
which touches the smaller circle C, at point P Na
B
To prove: Chord AB is bisected at point of contact (P}
i.e AP = BP
Solution:
Since AB is tangent to smaller circle C,
OP 1 AB (Tangent at any point of circle is perpendicular
to the radius through point of contact)
Example 2
Example 2 teachoo.com
Two tangents TP and TQ are drawn to a circle with centre O from an
external point T. Prove that ZPTQ = 2 ZOPQ
Pp
<h
Given: A circle with centre O
Two tangents TP and TQ to the circle <<
where P and Q are the point of contact <
Q
To prove: Z PTQ= 2 ZOPQ
Proof:
We know from theorem 10.2 that
length of tangents drawn from an external point to a circle are equal
So, TP=TQ
- ZTQP=ZTPQ {Angles opposite to equal sides are equal) ...(1)
Example 3
Example 3{Method 1) teachoo.com
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents
at P and Q intersect at a point T (see figure). Find the length TP.
Pp
Join OT.
Let OT intersect PQ at R Sem Sem
T ne R ae oO
From theorem 10.2,
Lengths of tangents from external point are equal @
So, TP=TQ
In ATPQ,
TP = TQ, i.e. two sides are equal,
So, A TPQ is an isosceles triangle
Here, OT is bisector of 2 PTQ,
So OT L PQ (Angle bisector and altitude
of isosceles triangle are same)
Therorems
2 questionsTheorem 10.1
Theorem 10.1 teachoo.com
The tangent at any point of a circle is perpendicular to the radius
through the point of contact.
Given: A circle with center O.
With tangent XY at point of contact P. o
To prove: OP L XY “Rx
P Q
Proof: Let Q be point on XY
Connect OQ
Suppose it touches the circle at R
Hence,
0Qa>OR
OQ > OP (as OP = OR radius )
Same will be the case with all other points on circle
Hence, OP is the smallest line that connects XY
Theorem 10.2
é
Theorem 10.2 (Method 1) sachoo.com
The lengths of tangents drawn from an external point to a circle are
equal.
Q
Given: Let circle be with centre O
and P bea point outside circle P
PQ and PR are two tangents to circle
intersecting at point Q and R respectively R
To prove: Lengths of tangents are equal
i.e. PQ=PR
Construction: Join OQ, OR and OP
Proof: As PQ is a tangent
OQ 1 PQ (Tangent at any point of circle is perpendicular
to the radius through point of contact)
So, Z OOP = 90°
Hence A OQP is right triangle
Case Based Questions (MCQ)
2 questionsQuestion 1
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods)
attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity. After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride . She was curious about the different angles
and measures that the wheel will form. She forms the figure as given below.
Question 1
In the given figure find ∠ROQ
(a) 60
(b) 100
(c) 150
(d) 90
Question 2
Find ∠RQP
(a) 75
(b) 60
(c) 30
(d) 90
Question 3
Find ∠RSQ
(a) 60
(b) 75
(c) 100
(d) 30
Question 4
Find ∠ORP
(a) 90
(b) 70
(c) 100
(d) 60
Question 2
Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff. The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O isinscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.
Question 1
Find the length of AD
(a) 7
(b) 8
(c) 5
(d) 9
Question 2
Find the Length of BE
(a) 8
(b) 5
(c) 2
(d) 9
Question 3
Find the length of CF
(a) 9
(b) 5
(c) 2
(d) 3
Question 4
If radius of the circle is 4cm, Find the area of ∆OAB
(a) 20
(b) 36
(c) 24
(d) 48
Question 5
Find area of ∆ABC
(a) 50
(b) 60
(c) 100
(d) 90
Why Learn This With Teachoo?
Circles is Chapter 10 of NCERT Class 10 Mathematics. It focuses on tangents: the number of tangents from different points, perpendicularity of radius and tangent, and equality of tangent lengths from an external point. Teachoo provides Exercises 10.1 and 10.2, examples, theorem proofs, numerical questions and case-based MCQs.
Tangents and points relative to a circle
A tangent touches a circle at exactly one point. A point inside the circle has no real tangent, a point on the circle has one and an external point has two.
The radius drawn to the point of contact is perpendicular to the tangent. This creates a right angle and enables Pythagoras-based length calculations.
From an external point P, tangent segments PA and PB to the same circle are equal. The proof joins the centre to A, B and P, creating right triangles with equal radii and common hypotenuse; RHS congruence establishes PA = PB.
These theorems combine with angle sums, quadrilateral properties and algebra in multi-step questions.
Topics available on Teachoo
-
Exercises 10.1 and 10.2 and examples;
-
number of tangents from a point;
-
tangent perpendicular to radius theorem;
-
numerical and proof questions for Theorem 10.1;
-
equal tangents from an external point;
-
numerical and proof questions for Theorem 10.2;
-
theorem practice and case-based questions.
Learning outcomes
Students should be able to determine the possible tangents from a point, use the radius–tangent right angle and apply equal tangent lengths. They should prove both core theorems and combine them with congruence, Pythagoras and angle properties.
Why is this chapter important?
Tangent geometry supports constructions, coordinate geometry and calculus ideas in later study. For boards, it is a compact but proof-heavy chapter where labelled diagrams and theorem statements matter.
How Teachoo helps
Teachoo separates proof and numerical forms. Mark the point of contact and draw radii immediately. Look for equal tangent pairs from the same external point and use them before introducing variables. In proofs, state the right angles, equal radii and congruence criterion.
Important concept connections
Tangent problems combine circle definitions with triangle congruence, Pythagoras and quadrilateral angle sums. The perpendicular radius creates the right triangle; equal radii and a common hypotenuse support RHS congruence; equal tangent lengths then simplify perimeters and algebraic expressions. The same perpendicularity theorem justifies the tangent construction in Chapter 11, so proof and construction should be studied together.
Board-exam and competency preparation
Circle questions usually require a short chain connecting tangents, radii and angles. Extend the diagram only with useful lines: radii to contact points and the segment joining the external point to the centre. This creates right triangles and exposes congruence or Pythagoras.
For a quadrilateral formed by two radii and two tangents, use the two right angles when applying the angle sum. If several external points appear, group tangent lengths by their common starting point. Proof questions need a valid congruence criterion, not the statement “tangents look equal.” Case-based questions may use circular parks or pulley diagrams; ignore decoration and identify centres, contact points and tangent segments.
Quick revision checklist
Count tangents for points inside, on and outside a circle; prove both core theorems; solve tangent-length and angle questions; and complete one diagram containing multiple external tangent pairs. Mark all right angles before calculating.
Common mistakes to avoid
A tangent is perpendicular to the radius only at the point of contact. Tangents from different external points are not automatically equal. Do not assume a secant is a tangent merely because the sketch appears to touch once.
Deeper reasoning and concept connections
The strongest way to learn Circles is to separate three layers: the object being studied, the rule that describes it and the reason the rule works. A correct numerical result is useful, but a complete mathematical answer also explains the relationship used. Students should compare examples and non-examples, change one condition at a time and observe whether the conclusion still holds.
This chapter is part of a longer progression. Its vocabulary and representations will appear again in algebra, geometry, data, measurement or higher problem-solving. Build links deliberately: translate pictures into statements, statements into operations and operations back into a sensible interpretation. If the final result cannot be explained in ordinary language, the method has probably been followed mechanically rather than understood.
How to solve unfamiliar and competency-based questions
When a question looks new, do not search memory for an identical example. Classify it. Decide whether it asks for recognition, calculation, representation, comparison, explanation or proof. Write the relevant definition or property first. Next, organise the data and select the shortest valid method. This converts an unfamiliar surface story into a familiar mathematical structure.
Use estimation and special cases as quality checks. Test zero, one, equal values, endpoints or a simple symmetric figure whenever they are permitted. A result that violates the diagram, scale, sign, unit or expected range is a signal to recheck the setup. In multi-part cases, carry forward only verified results so one early error does not silently contaminate every later answer.
What complete mastery looks like
For Circles, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Circles?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Circles?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
How many tangents can be drawn from an external point?
Two tangents can be drawn to a circle from an external point.
What angle does a tangent make with the radius at contact?
It makes a right angle, 90°.
Why are two tangents from the same external point equal?
They are corresponding sides of congruent right triangles formed with the centre and the external point.
Draw the radii to contact points first. Those two lines reveal nearly every useful relationship in the chapter.