CBSE Class 10 Maths Circles Competency Based Questions
Help your child master CBSE Class 10 Maths Circles important questions with our expert-verified competency based questions (CBQs). Covering tangents, concentric circles, case studies, and Assertion–Reason questions aligned with the latest CBSE board exam pattern — all with detailed step-by-step solutions prepared by Angle Belearn’s CBSE specialists.
CBSE Class 10 Maths Circles — Questions with Solutions
Two circles with centres O and N touch each other at point P as shown $\mathrm{O}, \mathrm{P}$ and N . The radius of the circle with centre O is twice that of the circle with centre $\mathrm{N} . \mathrm{OX}$ is a tangent to the circle with centre N , and OX $=18 \mathrm{~cm}$. What is the radius of the circle with centre N?
Option (C) is correct
Let $P N=x$ and $O P=2 x$ Join N with X and O , such that $\angle \mathrm{OXN}=90^{\circ}$, making $\Delta \mathrm{XON}$ as the right angled triangle. According to the Pythagoras theorem,
$$ \begin{array}{ll} \Rightarrow & O N^{2}=O N^{2}+O X^{2} \\ \Rightarrow & O N^{2}=x^{2}+18^{2} \\ \Rightarrow & (3 x)^{2}=x^{2}+324 \\ \Rightarrow & {[O N=O P+N P=2 x+x=3 x]} \\ \Rightarrow & 9 x^{2}=x^{2}+324 \\ & 8 x^{2}=324 \\ \Rightarrow & x^{2}=\frac{324}{8} \\ \Rightarrow & x=\sqrt{\frac{324}{8}} \\ & x=\sqrt{\frac{81}{2}}=\frac{9}{\sqrt{2}} \mathrm{~cm} \end{array} $$
$ QP $ is a tangent to a circle with centre $ O $ at a point $ P $ on the circle. If $ \triangle OPQ $ is isosceles, then $ \angle OQR $ equals:
Ans:
Let $ O $ be the centre of the circle. As per the given information, we have drawn the figure below.
.png)We know that the radius and tangent are perpendicular at their point of contact.
Now, in isosceles triangle $ POQ $, we have
$$\angle POQ + \angle OPQ + \angle OQP = 180^\circ$$
Equal sides subtend equal angles in isosceles triangle. Thus,
$$2 \times \angle OQP + 90^\circ = 180^\circ$$
$$\angle OQP = 45^\circ$$
Thus, $\angle OQR = 45^\circ$.
If a regular hexagon is inscribed in a circle of radius $ r $, then its perimeter is:
Side of the regular hexagon inscribed in a circle of radius $ r $ is also $ r $, therefore the perimeter is $ 6r $.
Thus, (b) is the correct option.
$ AB $ and $ CD $ are two common tangents to circles which touch each other at a point $ C $. If $ D $ lies on $ AB $ such that $ CD = 4 \, \text{cm} $, then $ AB $ is:
$ AD = CD $ and $ BD = CD $
$$AB = AD + BD = CD + CD = 2CD = 2 \times 4 = 8 \, \text{cm}$$
.png)Thus, (b) is the correct option.
The distance between two parallel tangents of a circle of diameter 7 cm is:
Given, Diameter of Circle = 7 cm
$$\text{Distance between two parallel tangents} = \text{diameter of circle} = 7 \, \text{cm}$$
Thus, (a) is the correct option.
Two parallel tangents are drawn to a circle at a distance of 10 cm, then the radius of the circle is:
Here, diameter $ d $ = Distance between two parallel tangents = 10 cm
.png)Thus, radius $ r = \frac{d}{2} = \frac{10}{2} = 5 \, \text{cm} $
Thus, (c) is the correct option.
In the given figure, $ QR $ is a common tangent to the given circle. Tangent at $ T $ meets $ QR $ at $ P $. If $ PQ = 5.5 \, \text{cm} $, then the length of $ QR $ is:
Given $ PQ = 5.5 \, \text{cm} $
The lengths of tangents drawn from an external point to a circle are equal. Thus,
$$PT = PQ = 5.5 \, \text{cm}$$
Again, $ P $ is an external point to a smaller circle. Therefore,
$$PR = PT = 5.5 \, \text{cm}$$
Now, the length of tangent $ QR = PQ + PR = 5.5 + 5.5 = 11 \, \text{cm}$
Thus, (c) is the correct option.
The length of tangent drawn to a circle of radius 9 cm from a point 41 cm from the centre is:
Given radius $ OA = 9 \, \text{cm} $ and $ OB = 41 \, \text{cm} $
.png)Since tangent is perpendicular to radius at point of contact, $ \angle OAB = 90^\circ $
In right-angled $ \triangle OAB $:
$$AB = \sqrt{OB^2 – OA^2} = \sqrt{41^2 – 9^2} = \sqrt{1681 – 81} = \sqrt{1600} = 40 \, \text{cm}$$
Thus, (a) is the correct option.
In the given figure, $ O $ is the centre of a circle. $ AB $ is a chord and the tangent $ AT $ at $ A $ makes an angle of 30° with the chord, then $ \angle OAB $ is:
.png)In the given figure,
$$\angle OAT = 90^\circ$$
$$\Rightarrow \angle OAB + \angle BAT = 90^\circ$$
$$\Rightarrow \angle OAB + 30^\circ = 90^\circ$$
$$\Rightarrow \angle OAB = 60^\circ$$
Thus, (c) is the correct option.
In figure, 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:
.png)Here, $ \angle OPT = \angle OQT = 90^\circ $, Given $ \angle POQ = 110^\circ $
In quadrilateral TPOQ:
$$\angle POQ + \angle OQT + \angle QTP + \angle TPO = 360^\circ$$
$$110^\circ + 90^\circ + \angle QTP + 90^\circ = 360^\circ$$
$$\Rightarrow \angle PTQ = 360^\circ – 290^\circ = 70^\circ$$
Read the following and answer any four questions from (i) to (v).
A Ferris wheel is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components attached to the rim. 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.
.png) 
.png)(i) In the given figure, $\angle ROQ$ is equal to:
(a) $60^\circ$ (b) $100^\circ$ (c) $150^\circ$ (d) $90^\circ$
(ii) The measurement of $\angle RQP$ is:
(a) $75^\circ$ (b) $60^\circ$ (c) $30^\circ$ (d) $90^\circ$
(iii) The measurement of $\angle RSQ$ is:
(a) $60^\circ$ (b) $75^\circ$ (c) $100^\circ$ (d) $30^\circ$
(iv) The measurement of $\angle ORP$ is:
(a) $90^\circ$ (b) $70^\circ$ (c) $100^\circ$ (d) $60^\circ$
(v) Reflex angle of $\angle ROQ$ is:
(a) $180^\circ$ (b) $150^\circ$ (c) $210^\circ$ (d) $360^\circ$
(i) In quadrilateral PROQ: $30^\circ + 90^\circ + \angle ROQ + 90^\circ = 360^\circ$
$\Rightarrow \angle ROQ = 150^\circ$ — Option (c)
(ii) In $\triangle OQR$, $OQ = OR$ (radii), so $\angle OQR = \angle ORQ$.
$150^\circ + 2\angle OQR = 180^\circ \Rightarrow \angle OQR = 15^\circ$
$\angle RQP = 90^\circ – 15^\circ = 75^\circ$ — Option (a)
(iii) $\angle RSQ = \frac{1}{2}\angle QOR = \frac{1}{2} \times 150^\circ = 75^\circ$ — Option (b)
(iv) $OR \perp PR$ (radius ⊥ tangent) $\Rightarrow \angle ORP = 90^\circ$ — Option (a)
(v) Reflex $\angle ROQ = 360^\circ – 150^\circ = 210^\circ$ — Option (c)
Read the following and answer any four questions from (i) to (v).
Varun has been selected by his School to design the logo for Sports Day T-shirts. In the given figure, a circle with centre $O$ is inscribed in a $\triangle ABC$, 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.
.png)(i) The length of $AD$ is:
(a) 7 cm (b) 8 cm (c) 5 cm (d) 9 cm
(ii) The length of $BE$ is:
(a) 8 cm (b) 5 cm (c) 2 cm (d) 9 cm
(iii) The length of $CF$ is:
(a) 9 cm (b) 5 cm (c) 2 cm (d) 3 cm
(iv) If radius of the circle is 4 cm, then the area of $\triangle OAB$ is:
(a) $20 \, \text{cm}^2$ (b) $36 \, \text{cm}^2$ (c) $24 \, \text{cm}^2$ (d) $48 \, \text{cm}^2$
(v) The area of $\triangle ABC$ is:
(a) $50 \, \text{cm}^2$ (b) $60 \, \text{cm}^2$ (c) $100 \, \text{cm}^2$ (d) $90 \, \text{cm}^2$
.png)(i) Let $AD = x$. Then $DB = 12 – x$, $BE = 12 – x$, $CE = x – 4$, $CF = x – 4$, $AF = 14 – x$.
Since $AF = AD$: $14 – x = x \Rightarrow x = 7$ cm — Option (a)
(ii) $BE = 12 – 7 = 5$ cm — Option (b)
(iii) $CF = 7 – 4 = 3$ cm — Option (d)
(iv) Area of $\triangle OAB = \frac{1}{2} \times AB \times OD = \frac{1}{2} \times 12 \times 4 = 24 \, \text{cm}^2$ — Option (c)
(v) Area of $\triangle ABC = 24 + \frac{1}{2}(8)(4) + \frac{1}{2}(10)(4) = 24 + 16 + 20 = 60 \, \text{cm}^2$ — Option (b)
If a tangent is drawn to a circle from an external point, then the radius at the point of contact is perpendicular to the tangent.
Based on the above information, solve the following questions:
Q1. In the given figure, $O$ is the centre of two concentric circles. From an external point $P$, tangents $PA$ and $PB$ are drawn such that $PA = 6$ cm and $PB = 8$ cm. If $OP = 10$ cm, find the value of $AB$.
.png)Q2. The diameters of two concentric circles are 10 cm and 6 cm. $AB$ is a diameter of the bigger circle and $BD$ is tangent to the smaller circle touching it at $D$ and intersecting the larger circle at $P$ on producing. Find the length of $BP$.
.png)Q3. Two concentric circles are such that the difference between their radii is 4 cm and the length of the chord of the larger circle which touches the smaller circle is 24 cm. Find the radius of the smaller circle.
Or: If $AB$ is a chord of a circle with centre $O$, $AOC$ is a diameter and $AT$ is the tangent at $A$, prove that $\angle BAT = \angle ACB$.
.png)Q1. Since radius ⊥ tangent: $OB \perp BP$ and $OA \perp AP$.
$$OB = \sqrt{OP^2 – PB^2} = \sqrt{100 – 64} = 6 \, \text{cm}$$
$$OA = \sqrt{OP^2 – PA^2} = \sqrt{100 – 36} = 8 \, \text{cm}$$
$$AB = OA – OB = 8 – 6 = 2 \, \text{cm}$$
Q2. Since radius ⊥ tangent: $OD \perp BP$. $OB = 5$ cm, $OD = 3$ cm.
$$BD = \sqrt{OB^2 – OD^2} = \sqrt{25 – 9} = 4 \, \text{cm}$$
Since chord $BP$ is bisected by radius $OD$:
$$BP = 2 \times BD = 8 \, \text{cm}$$
Q3. Let $x$ = radius of smaller circle, $(x+4)$ = radius of larger circle.
$$OD \perp AB, \; AD = 12 \, \text{cm}$$
$$(x + 4)^2 = x^2 + 12^2 \Rightarrow 8x + 16 = 144 \Rightarrow x = 16 \, \text{cm}$$
.png)Or: Since $AC$ is a diameter, $\angle ABC = 90^\circ$ (angle in semicircle).
In $\triangle ABC$: $\angle CAB + \angle ACB = 90^\circ$ …(1)
Since $CA \perp AT$: $\angle CAB + \angle BAT = 90^\circ$ …(2)
From (1) and (2): $\angle ACB = \angle BAT$. Hence proved.
Smita always finds it confusing with the concepts of tangent and secant of a circle. Here, some questions are listed to clear your concepts.
(i) A line that intersects a circle exactly at two points is called:
(a) Secant (b) Tangent (c) Chord (d) Both (a) and (b)
(ii) Number of tangents that can be drawn to a circle is:
(a) 1 (b) 0 (c) 2 (d) Infinite
(iii) Number of tangents that can be drawn to a circle from a point not on it is:
(a) 1 (b) 2 (c) 0 (d) Infinite
(iv) Number of secants that can be drawn to a circle from a point on it is:
(a) Infinite (b) 1 (c) 2 (d) 0
(v) A line that touches a circle at only one point is called:
(a) Secant (b) Chord (c) Tangent (d) Diameter
(i) Answer: (a) Secant — A secant intersects a circle at exactly two points.
(ii) Answer: (d) Infinite — Infinitely many tangents can be drawn to a circle (one at every point on the circle).
(iii) Answer: (b) 2 — Exactly two tangents can be drawn from a point outside the circle.
(iv) Answer: (a) Infinite — Infinitely many secants can be drawn from a point on the circle.
(v) Answer: (c) Tangent — A tangent touches the circle at exactly one point.
A backyard is in the shape of a triangle with right angle at $B$, $AB = 6$ m and $BC = 8$ m. A circular pit was dug inside it such that it touches the walls $AC$, $BC$, and $AB$ at $P$, $Q$, and $R$ respectively, with $AP = x$ m.
Based on the above information, answer the following questions:
(i) The value of $AR =$
(a) $2x$ m (b) $x/2$ m (c) $x$ m (d) $3x$ m
(ii) The value of $BQ =$
(a) $2x$ m (b) $(6 – x)$ m (c) $(2 – x)$ m (d) $4x$ m
(iii) The value of $CQ =$
(a) $(4 + x)$ m (b) $(10 – x)$ m (c) $(2 + x)$ m (d) both (b) and (c)
(iv) Which of the following is correct?
(a) Quadrilateral $AROF$ is a square. (b) Quadrilateral $BROQ$ is a square.
(c) Quadrilateral $CQOP$ is a square. (d) None of these.
(v) Radius of the pit is:
(a) 2 cm (b) 3 cm (c) 4 cm (d) 5 cm
(i) (c) $AR = x$ m (tangent lengths from A are equal: $AR = AP = x$)
(ii) (b) $BQ = (6 – x)$ m
(iii) (d) both (b) and (c)
(iv) (b) Quadrilateral $BROQ$ is a square.
(v) (b) Radius = 3 cm
*Using $AC = \sqrt{AB^2 + BC^2} = 10$ m. Setting up equations with tangent lengths gives $x = 3$ m and radius $r = 6 – x = 3$ cm.*
From an external point $ Q $, the length of tangent to a circle is 12 cm and the distance of $ Q $ from the centre of the circle is 13 cm. The radius of the circle (in cm) is:
Given $ OQ = 13 \, \text{cm} $ and $ PQ = 12 \, \text{cm} $. Since radius ⊥ tangent, $ OP \perp PQ $.
$$OP^2 = OQ^2 – PQ^2 = 169 – 144 = 25 \Rightarrow OP = 5 \, \text{cm}$$
Thus, (b) is the correct option.
A chord of a circle of radius 10 cm, subtends a right angle at its centre. The length of the chord (in cm) is:
.png)Using Pythagoras theorem in $ \triangle ABC $:
$$BC^2 = AB^2 + AC^2 = 10^2 + 10^2 = 200$$
$$BC = 10\sqrt{2} \, \text{cm}$$
Thus, (c) is the correct option.
In figure, O is the centre of the circle. PQ is a chord and PT is tangent at P which makes an angle of $ 50^\circ $ with PQ. Then $ \angle POQ $ is:
.png)Since radius ⊥ tangent: $ \angle OPT = 90^\circ $
$$\angle OPQ = 90^\circ – 50^\circ = 40^\circ$$
Since $ OP = OQ $ (radii): $ \angle OPQ = \angle OQP = 40^\circ $
$$\angle POQ = 180^\circ – 40^\circ – 40^\circ = 100^\circ$$
Thus, (c) is the correct option.
In the given figure, the perimeter of $ \triangle ABC $ is:
.png)$ AR = AQ = 5 \, \text{cm} $, $ BP = BR = 6 \, \text{cm} $, $ CQ = CP = 4 \, \text{cm} $
$$\text{Perimeter} = (5+6) + (6+4) + (4+5) = 11 + 10 + 9 = 30 \, \text{cm}$$
A quadrilateral PQRS is drawn to circumscribe a circle. If $ PQ = 12 \, \text{cm} $, $ QR = 15 \, \text{cm} $ and $ RS = 14 \, \text{cm} $, find the length of $ SP $.
For a quadrilateral circumscribing a circle:
$$PQ + RS = SP + QR$$
$$SP = PQ + RS – QR = 12 + 14 – 15 = 11 \, \text{cm}$$
.png)In the given figure, if AB and AC are two tangents to a circle with centre O, so that $ \angle BOC = 100^\circ $, then $ \angle OAB $ is:
.png)$$\angle BOC + \angle BAC = 180^\circ \Rightarrow \angle BAC = 80^\circ$$
Since OA bisects $ \angle BAC $:
$$\angle OAB = \frac{80^\circ}{2} = 40^\circ$$
If radii of two concentric circles are 6 cm and 4 cm, the length of the chord of the larger circle that touches the smaller circle is:
Given $ OA = 6 \, \text{cm} $ (larger) and $ OB = 4 \, \text{cm} $ (smaller).
.png)By Pythagoras theorem:
$$AB = \sqrt{OA^2 – OB^2} = \sqrt{36 – 16} = \sqrt{20} = 2\sqrt{5} \, \text{cm}$$
Full chord $ = 2 \times AB = 4\sqrt{5} \, \text{cm} $
In figure, $ AP $, $ AQ $ and $ BC $ are tangents of the circle with centre $ O $. If $ AB = 5 \, \text{cm} $, $ AC = 6 \, \text{cm} $ and $ BC = 4 \, \text{cm} $, then the length of $ AP $ (in cm) is:
.png)Since $ BP = BR $, $ CR = CQ $, $ AP = AQ $:
$$AB + BC + AC = AP + AQ = 2AP$$
$$2AP = 5 + 4 + 6 = 15 \Rightarrow AP = 7.5 \, \text{cm}$$
Two circles of radii 20 cm and 37 cm intersect at $ A $ and $ B $. If $ O_1 $ and $ O_2 $ are their centres and $ AB = 24 $ cm, then the distance $ O_1O_2 $ is equal to:
Let C be the midpoint of AB, so $ AC = 12 \, \text{cm} $. $ AO_1 = 37 \, \text{cm} $, $ AO_2 = 20 \, \text{cm} $.
.png)$$CO_1 = \sqrt{37^2 – 12^2} = \sqrt{1369 – 144} = 35 \, \text{cm}$$
$$CO_2 = \sqrt{20^2 – 12^2} = \sqrt{400 – 144} = 16 \, \text{cm}$$
$$O_1O_2 = CO_1 + CO_2 = 35 + 16 = 51 \, \text{cm}$$
Two concentric circles have radii 5 cm and 13 cm. The length of the chord of the larger circle which touches the smaller circle is:
The chord of the larger circle is tangent to the smaller circle, so distance from centre to chord = radius of smaller circle = 5 cm.
$$\text{Half chord} = \sqrt{R^2 – d^2} = \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12 \, \text{cm}$$
$$\text{Full chord} = 2 \times 12 = 24 \, \text{cm}$$
In a park, four poles are standing at positions A, B, C, and D around the fountain such that the cloth joining the poles AB, BC, CD, and DA touches the fountain at P, Q, R, and S respectively.
.png) 
.png)Q1. If O is the centre of the circular fountain, then $\angle OSA =$
(a) $60^\circ$ (b) $90^\circ$ (c) $45^\circ$ (d) None of these
Q2. Which of the following is correct?
(a) $AS = AP$ (b) $BP = BQ$ (c) $CQ = CR$ (d) All of these
Q3. If $DR = 7$ cm and $AD = 11$ cm, then $AP =$
(a) 4 cm (b) 18 cm (c) 7 cm (d) 11 cm
Q4. If $\angle QCR = 60^\circ$, then $\angle QOR =$
(a) $60^\circ$ (b) $120^\circ$ (c) $90^\circ$ (d) $30^\circ$
Q5. Which of the following is correct?
(a) $AB + BC = CD + DA$ (b) $AB + AD = BC + CD$ (c) $AB + CD = AD + BC$ (d) All of these
Q1. (b) $\angle OSA = 90^\circ$ (radius ⊥ tangent at point of contact)
.png)Q2. (d) All of these — tangent lengths from an external point are equal:
$AS = AP$, $BP = BQ$, $CQ = CR$, $DR = DS$
Q3. (a) $AP = AS = AD – DS = AD – DR = 11 – 7 = 4$ cm
Q4. (b) In quadrilateral OQCR: $\angle QOR = 360^\circ – 90^\circ – 90^\circ – 60^\circ = 120^\circ$
.png)Q5. (c) $AB + CD = AD + BC$
For class 10 students, a teacher planned a game for the revision of Circles chapter. Based on the given information, solve the following questions:
Q1. In the given figure, $x + y =$
(a) $60^\circ$ (b) $90^\circ$ (c) $120^\circ$ (d) $145^\circ$
Q2. If PA and PB are two tangents from P with $\angle PBA = 50^\circ$, then $\angle OAB =$
(a) $50^\circ$ (b) $25^\circ$ (c) $40^\circ$ (d) $130^\circ$
Q3. In the given figure, PQ and PR are two tangents to the circle, then $\angle ROQ =$
(a) $30^\circ$ (b) $60^\circ$ (c) $105^\circ$ (d) $150^\circ$
Q4. In the given figure, AB is a chord, AOC is its diameter, $\angle ACB = 55^\circ$, then $\angle BAT =$
(a) $35^\circ$ (b) $55^\circ$ (c) $125^\circ$ (d) $110^\circ$
Q5. In the given figure, PC is the tangent at A, $\angle PAB = 72^\circ$ and $\angle AOB = 132^\circ$, then $\angle ABC =$
(a) $18^\circ$ (b) $30^\circ$ (c) $60^\circ$ (d) Can’t be determined
Q1. (b) $x + y = 90^\circ$ (radius ⊥ tangent: $\angle OAC + \angle OCA = 90^\circ$)
Q2. (c) $\angle OBA = 90^\circ – 50^\circ = 40^\circ$. Since $OA = OB$, $\angle OAB = \angle OBA = 40^\circ$
.png)Q3. (d) $\angle ROQ + \angle RPQ = 180^\circ \Rightarrow \angle ROQ = 180^\circ – 30^\circ = 150^\circ$
Q4. (b) $\angle ABC = 90^\circ$, $\angle BAC = 35^\circ$, $\angle BAT = 90^\circ – 35^\circ = 55^\circ$
Q5. (b) $\angle OAB = 18^\circ$, In $\triangle OAB$: $\angle ABO = 180^\circ – 18^\circ – 132^\circ = 30^\circ$, so $\angle ABC = 30^\circ$
In a math class, the teacher draws two circles that touch each other externally at point M with centres A and B and radii 5 cm and 4 cm respectively.
.png)Based on the above information, solve the following questions:
Q1. Find the value of $PX$.
Q2. Find the value of $QY$.
Q3. Show that $PS^2 = PM \cdot PX$.
Or Show that $TQ^2 = YQ \cdot MQ$.
Q1. In right-angled $\triangle ASP$ ($AS = 5$ cm, $PS = 12$ cm):
$$PA = \sqrt{12^2 + 5^2} = \sqrt{169} = 13 \, \text{cm}$$
$$PX = PA – XA = 13 – 5 = 8 \, \text{cm}$$
Q2. In right-angled $\triangle BTQ$ ($BT = 4$ cm, $TQ = 3$ cm):
$$BQ = \sqrt{3^2 + 4^2} = 5 \, \text{cm}$$
$$QY = BQ – BY = 5 – 4 = 1 \, \text{cm}$$
Q3. In $\triangle ASP$:
$$PS^2 = PA^2 – AM^2 = (PA + AM)(PA – AM) = (PA + AM)(PA – AX) = PM \cdot PX$$
Hence proved.
Or: $TQ^2 = (BQ^2 – BM^2) = (BQ – BM)(BQ + BM) = (BQ – BY) \cdot MQ = YQ \cdot MQ$. Hence proved.
If a tangent is drawn to a circle from an external point, then the radius at the point of contact is perpendicular to the tangent.
Based on the above information, solve the following questions:
Q1. O is the centre of two concentric circles. From external point P, tangents PA and PB are drawn such that $PA = 6$ cm, $PB = 8$ cm and $OP = 10$ cm. Find $AB$.
.png)Q2. Diameters of two concentric circles are 10 cm and 6 cm. AB is a diameter of the bigger circle and BD is tangent to the smaller circle touching it at D and intersecting the larger circle at P. Find BP.
.png)Q3. Two concentric circles are such that the difference between their radii is 4 cm and the chord of the larger circle touching the smaller circle is 24 cm. Find the radius of the smaller circle.
Or: AB is a chord, AOC is a diameter and AT is tangent at A. Prove $\angle BAT = \angle ACB$.
.png)Q1. $OB = \sqrt{100-64} = 6$ cm, $OA = \sqrt{100-36} = 8$ cm
$$AB = OA – OB = 2 \, \text{cm}$$
Q2. $OB = 5$ cm, $OD = 3$ cm
$$BD = \sqrt{25-9} = 4 \, \text{cm} \Rightarrow BP = 2 \times 4 = 8 \, \text{cm}$$
Q3. Let smaller radius $= x$, larger $= x + 4$. Half-chord $= 12$ cm.
$$(x+4)^2 = x^2 + 144 \Rightarrow 8x + 16 = 144 \Rightarrow x = 16 \, \text{cm}$$
Or: $\angle ABC = 90^\circ$ (angle in semicircle)
$$\angle CAB + \angle ACB = 90^\circ \quad \cdots (1)$$
$$CA \perp AT \Rightarrow \angle CAB + \angle BAT = 90^\circ \quad \cdots (2)$$
From (1) & (2): $\angle ACB = \angle BAT$. Hence proved.
Circles play an important part in our life. When a circular object is hung on the wall with a chord at nail N, the chords NA and NB work like tangents. Observe the figure, given that $\angle ANO = 30^\circ$ and $OA = 5$ cm.
.png)Based on the above information, solve the following questions:
Q1. Find the distance $AN$.
Q2. Find the measure of $\angle AOB$.
Q3. Find the total length of chords $NA$, $NB$ and the chord $AB$.
Or: Name the type of quadrilateral $OANB$ and justify your answer.
Q1. Since $\angle OAN = 90^\circ$ (radius ⊥ tangent) and $\angle ANO = 30^\circ$:
$$\tan 30^\circ = \frac{OA}{AN} \Rightarrow AN = \frac{OA}{\tan 30^\circ} = \frac{5}{1/\sqrt{3}} = 5\sqrt{3} \, \text{cm}$$
Q2. $\angle AON = 90^\circ – 30^\circ = 60^\circ$, so $\angle AOB = 2 \times 60^\circ = 120^\circ$
Q3. $NA = NB = 5\sqrt{3}$ cm (equal tangents). Chord $AB$: using $\triangle OAB$ with $\angle AOB = 120^\circ$ and $OA = OB = 5$ cm, $AB = 5\sqrt{3}$ cm.
Total $= 5\sqrt{3} + 5\sqrt{3} + 5\sqrt{3} = 15\sqrt{3}$ cm
Or: Quadrilateral OANB is a kite — $OA = OB$ (radii) and $NA = NB$ (equal tangents). The diagonal ON bisects AB perpendicularly.
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R): The lengths of tangents drawn from the external point to a circle are equal.
Answer: (B)
Both A and R are true. A is a theorem about perpendicularity of tangent and radius. R is a separate theorem about equal tangent lengths. R does not explain A, so (B) is correct.
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In the given figure, O is the centre of a circle and AT is a tangent at point A, then $\angle BAT = 70^\circ$.
.png)Reason (R): A straight line can intersect a circle at one point only.
Answer: (C)
A is true — by the alternate segment theorem, $\angle BAT = \angle ACB = 70^\circ$.
R is false — a straight line can intersect a circle at two points (that would be a secant).
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The distance between two parallel tangents of a circle is 16 cm, then the radius of the circle is 10 cm.
Reason (R): The distance between two parallel tangents of a circle is equal to the diameter of a circle.
Answer: (D)
R is true — distance between parallel tangents = diameter.
A is false — if diameter = 16 cm, then radius = 8 cm, not 10 cm.
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If PA and PB are tangents from external point P to a circle with centre O, then quadrilateral AOBP is cyclic.
Reason (R): The angle between 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: (A)
Both A and R are true. $\angle APB + \angle AOB = 180^\circ$ and $\angle OAP + \angle OBP = 180^\circ$, making AOBP a cyclic quadrilateral. Reason R directly explains this.
.png)Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a quadrilateral ABCD drawn to circumscribe a circle, $AB + BC = AD + DC$.
.png)Reason (R): In two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Answer: (D)
A is false — the correct relation for a circumscribed quadrilateral is $AB + CD = AD + BC$.
R is true — the perpendicular from the centre bisects the chord, so the chord is bisected at the point of tangency.
.png)Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): PA and PB are two tangents to a circle with centre O such that $\angle AOB = 110^\circ$, then $\angle APB = 90^\circ$.
Reason (R): The length of two tangents drawn from an external point are equal.
Answer: (D)
A is false: $\angle APB = 360^\circ – 90^\circ – 90^\circ – 110^\circ = 70^\circ$ (not $90^\circ$).
.png)R is true: tangent lengths from an external point are equal.
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If in a circle, radius = 3 cm and distance of a point from centre = 5 cm, then the length of the tangent will be 4 cm.
Reason (R): $(\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2$
Answer: (A)
Both true. $AB = \sqrt{OB^2 – OA^2} = \sqrt{25 – 9} = 4$ cm.
.png)The Pythagorean theorem (R) directly explains how to calculate the tangent length (A).
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): Two tangents are drawn to a circle from an external point; they subtend equal angles at the centre.
Reason (R): A parallelogram circumscribing a circle is a rhombus.
Answer: (B)
Both A and R are true. A is correct — tangents from an external point subtend equal angles at centre. R is also true. But R does not explain A.
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In a cyclic quadrilateral, one angle is $40^\circ$, then the opposite angle is $140^\circ$.
Reason (R): Sum of opposite angles in a cyclic quadrilateral is equal to $360^\circ$.
Answer: (C)
A is true: opposite angle $= 180^\circ – 40^\circ = 140^\circ$ (since opposite angles in a cyclic quadrilateral are supplementary).
R is false: the sum of opposite angles in a cyclic quadrilateral is $180^\circ$, not $360^\circ$.
Assertion–Reason (A–R)
Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): Centre and radius of the circle $x^2 + y^2 – 6x + 4y – 36 = 0$ is $(3, -2)$ and 7 respectively.
Reason (R): Centre and radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $(-g, -f)$ and $\sqrt{g^2 + f^2 – c}$ respectively.
Answer: (A)
$2g = -6 \Rightarrow g = -3$, $2f = 4 \Rightarrow f = 2$
Centre $= (-g, -f) = (3, -2)$ ✓
$$r = \sqrt{g^2 + f^2 – c} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$ ✓
R is the standard formula that explains why A is correct.
