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Maths M.C.Q (1 Marks)

Circles · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 59 questions.

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M.C.Q (1 Marks)

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59 questions · auto-graded multiple-choice test.

MCQ 11 Mark
In the given figure, $RQ$ is a tangent to the circle with centre $O$. If $SQ = 6\ cm$ and $QR = 4\ cm$, then $OR$ is equal to:
  • A
    $2.5\ cm$
  • B
    $3\ cm$
  • ✓
    $5\ cm$
  • D
    $8\ cm$
Answer
Correct option: C.
$5\ cm$
$SQ=6 \ cm \Rightarrow OQ=3 \ cm$
$QR=4 \ cm$
Since $R Q$ is a tangent to the circle at $Q$.
$\angle RQO =90^{\circ}...($tangent is perpendicular to the radius of a circle$)$
$\text { In } \triangle RQO$
By using Pythagoras theorem,
$O R^2=R Q^2+O Q^2$
$=4^2+3^2$
$=16+9$
$=25$
$\therefore O R^2=25$
$\Rightarrow O R=5 \ cm$
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MCQ 21 Mark
Each question consists of two statements, namely, Assertion $(A)$ and Reason $(R).$ for selecting the correct answer, use the following code:
Assertion $(A)$
Reason $(R)$
At a point $P$ of a circle with centre $O$ and radius $12\ cm,$ a tangent $PQ$ of length $16\ cm$ is drawn. Then, the point of contact. $OQ = 20\ cm.$
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  • ✓
    Both Assertion $(A)$ and Reason $(R)$ are true and Reason $(R)$ is a correct explanation of Assertion $(A)$.
  • B
    Both Assertion $(A)$ and Reason $(R)$ are true but Reason $(R)$ is not a correct explanation of Assertion $(A).$
  • C
    Assertion $(A)$ is true and Reason $(R)$ is false.
  • D
    Assertion $(A)$ is false and Reason $(R)$ is true.
Answer
Correct option: A.
Both Assertion $(A)$ and Reason $(R)$ are true and Reason $(R)$ is a correct explanation of Assertion $(A)$.

We know that the tangent is perpendicular to the radius of a circle.
In $\triangle\text{OPQ},$
By Pythagoras theorem,
$OQ^2= QP^2 + OP^2$
$\Rightarrow OP^2 = 16^2 + 12^2$
$\Rightarrow OP^2 = 256 + 144$
$\Rightarrow OP^2 = 400$
$\Rightarrow OP = 20\ cm$
So, the Assertion $(A)$ is true.
The Reason $(R)$ is true and is the correct explanation for the Assertion $(A).$
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MCQ 31 Mark
In the given figure, QR is a common tangent to the given circle, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8cm then the length of QR is:
  • A
    1.9cm
  • B
    3.8cm
  • C
    5.7cm
  • ✓
    7.6cm
Answer
Correct option: D.
7.6cm
We know that tangent from an external point to the circle are equal.
PQ = PT = 3.8cm
PR = PT = 3.8cm
QR = PQ + PR
= 3.8 +3.8
=7.6cm
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MCQ 41 Mark
In the given figure, AB and AC are tangent to the circle with centre O such that $\angle\text{BAC}=40^\circ.$ Then, $\angle\text{BOC}$ is equal to:
  • A
    80º
  • B
    100º
  • C
    120º
  • ✓
    140º
Answer
Correct option: D.
140º
Since AB and AC are the tangent to the circle.
$\angle\text{OBA}=\angle\text{BAC}=90^\circ$....(tangent is perpendicular to the radius of a circle)
In ABOC
$\angle\text{OBA}+\angle\text{BAC}+\angle\text{OCA}+\angle\text{BOC}=360^\circ$
$\Rightarrow90^\circ+40^\circ+90^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=140^\circ$
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MCQ 51 Mark
In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5cm, BC = 7cm and CS = 3cm, AB =?
  • ✓
    9cm
  • B
    10cm
  • C
    12cm
  • D
    8cm
Answer
Correct option: A.
9cm
We know that tangent from an external point to the circle are equal.
AP = AQ = 5cm
CS = CR = 3cm
RB = BC - CR
= 7 +3
=4cm
So, BQ = RB = 4cm
Thus, AB = AQ + RB = 5 + 4 = 9cm
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MCQ 61 Mark
In the given figure, $O$ is the centre of two concentric circles of radii $5\ cm$ and $3\ cm.$ From an external point $P$ tangents $PA$ and $PB$ are drawn to these circles. If $PA = 12\ cm$ then $PB$ is equal to:
  • A
    $5\sqrt{2}\text{cm}$
  • B
    $3\sqrt{5}\text{cm}$
  • ✓
    $4\sqrt{10}\text{cm}$
  • D
    $5\sqrt{10}\text{cm}$
Answer
Correct option: C.
$4\sqrt{10}\text{cm}$
Construction: Join $OB.$
We know that tangent is perpendicular to the radius of a circle.
In $\triangle\text{OPA},$
By Pythagoras theorem,
$OP^2 = OA^2 + AP^2$
$\Rightarrow OP^2 = 5^2 + 12^2$
$\Rightarrow OP^2 = 169$
$\Rightarrow OP = 13cm$
In $\triangle\text{OPB},$
By Pythagoras theorem,
$OP^2 = OB^2 + PB^2$
$\Rightarrow PB^2= OP^2- OB^2$
$\Rightarrow PB^2 = 13^2 - 3^2$
$\Rightarrow PB^2= 160$
$\Rightarrow \text{PB} =4\sqrt{10}\text{cm} $
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MCQ 71 Mark
In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If $\angle\text{PAO} = 30^\circ$ then $\angle\text{CPB} + \angle\text{ACP}$ is equal to:
  • A
    60º
  • ✓
    90º
  • C
    120º
  • D
    150º
Answer
Correct option: B.
90º
Since APB is a straight line,$\angle\text{APD}+\angle\text{DPC}+\angle\text{CPB}=180º$
We know that angles that subtend the same arc are equal.
$\Rightarrow\angle\text{APD}=\angle\text{ACP}$
$\Rightarrow\angle\text{ACP}+90^\circ+\angle\text{CPB}=180^\circ$....(Since $\angle\text{DPC}$ is inscribed in a semicircle)
$\Rightarrow\angle\text{CPB}+\angle\text{ACP}=90^\circ$
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MCQ 81 Mark
Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:
Assertion (A)
Reason (R)
If two tangent are drawn to a circle from an external point then they subtend equal angles at the centre. A parallelogram circumscribing a circle is a rhombus.
  • A
    Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
  • ✓
    Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
  • C
    Assertion (A) is true and Reason (R) is false.
  • D
    Assertion (A) is false and Reason (R)is true.
Answer
Correct option: B.
Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).


Consider tangent AB and AC drawn to the circle with centre O.
In $​​\triangle\text{OBA}$ and $​​\triangle\text{OCA},$
$\text{AO}=\text{AO}$ ....(common side)
$\text{OB}=\text{OC}$ .....(radii of the same circle)
$\angle\text{B}=\angle\text{C}=90^\circ$
$\Rightarrow\triangle\text{OBA}\cong\triangle\text{OCA}$ ....(RHS congruence criterion)
So, $\angle\text{OBA}=\angle\text{COA}$ ....(cpct)
Thus, the (R) is also true and can be proved using the property, 'tangent from an external point to a circle are equal'
But, the Reason (R) is not the correct explanation for the Assertion (A).
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MCQ 91 Mark
In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A. If $\angle\text{AOB} = 100^\circ$ then $\angle\text{BAT} $ is equal to:
  • A
    40º
  • ✓
    50º
  • C
    90º
  • D
    100º
Answer
Correct option: B.
50º
In $\triangle\text{OAB},$$\text{OA}=\text{OB}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OAB}=\angle\text{OAB}$ ....(angles opposite equal sides are equal)
$\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$ (Angle Sum Property)
$\Rightarrow\angle\text{AOB}+\angle\text{OAB}+\angle\text{OAB}=180^\circ$
$\Rightarrow100^\circ+2\angle\text{OAB}=180^\circ$
$\Rightarrow2\angle\text{OAB}=80^\circ$
$\Rightarrow\angle\text{OAB}=40^\circ$
Since AT is the tangent,
$\angle\text{OAT}=90^\circ$
$\Rightarrow\angle\text{OAB}+\angle\text{BAT}=90^\circ$
$\Rightarrow40^\circ+\angle\text{BAT}=90^\circ$
$\Rightarrow\angle\text{BAT}=50^\circ$
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MCQ 101 Mark
In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively, and S is a point on the circle, such that $\angle\text{SQL} = 50^\circ.$ and $\angle\text{SRM} = 60^\circ.$ Find $ \angle\text{QSR}=?$
  • A
    40º
  • B
    50º
  • C
    60º
  • ✓
    70º
Answer
Correct option: D.
70º
Since PL and PM are the tangent to the circle.$\angle\text{OQL}=\angle\text{ORM}=90^\circ$ (tangent is perpendicular to the radius of a circle)
So,
$\angle\text{OQL}+\angle\text{SQL}+\angle\text{OQS}$
$\Rightarrow90^\circ=50^\circ+\angle\text{OQS}$
$\Rightarrow\angle\text{OQC}=40^\circ$
Similarly, we can find $\angle\text{ORS}=30^\circ.$
In $\triangle\text{OQS},$
$\text{OQ}=\text{OS}$
$\angle\text{OQS}=\angle\text{OSQ}=40^\circ$ ...(angles opposite equal sides are equal)
In $\triangle\text{ORS},$
$\text{OR}=\text{OS}$
$\angle\text{ORS}=\angle\text{OSR}=30^\circ$
So, $\angle\text{QSR}=\angle\text{OSQ}+\angle\text{OSR}=40^\circ+30^\circ=70^\circ.$
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MCQ 111 Mark
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5cm, AC = 6cm and BC = 4cm then the length of AP is:
  • A
    15cm
  • B
    10cm
  • C
    9cm
  • ✓
    7.5cm
Answer
Correct option: D.
7.5cm
Let BC intersect the circle at D.We know that tangent from an external point to the circle are equal.
BP = BD
CD = CQ
AP = AQ
perimeter of $\triangle\text{ABC}$
= AB + BC + AC
= AB + (BD + CD) + AC
= AB + (BP + CQ) + AC
= (AB + BP) + (AC + CQ)
= AP + AQ
Since perimeter of $\triangle\text{ABC}$ = AB + BC + AC = 5 + 6 + 4 = 15cm
⇒ AP + AQ = 15
⇒ 2AP = 15
⇒ AP = 7.5cm
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MCQ 121 Mark
In the given figure, $PT$ is a tangent to a circle with centre $O$. If $OT = 6\ cm$ and $OP = 10\ cm,$ then the length of tangent $PT$ is:
  • ✓
    $8\ cm$
  • B
    $10\ cm$
  • C
    $12\ cm$
  • D
    $16\ cm$
Answer
Correct option: A.
$8\ cm$
$OT = 6\ cm$
$OP = 10\ cm$
Since $PT$ is a tangent to the circle at $T.$
$\angle\text{PTO}=90^\circ....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{PTO},$
By Pythagoras theorem,
$OP2 = PT^2 + OT^2$
$\Rightarrow PT^2 = OP^2 - OT^2$
$\Rightarrow PT^2 = 10^2- 6^2$
$\Rightarrow PT^2 = 100 - 36$
$\Rightarrow PT = 8\ cm$
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MCQ 131 Mark
In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10cm, BC = 38cm, PB = 27cm and $\text{AD} \perp \text{CD}$ then the length of CD is:
  • A
    11cm
  • B
    15cm
  • C
    20cm
  • ✓
    21cm
Answer
Correct option: D.
21cm
We know that tangles from an external point to the circle are equal.
BQ = PB = 27cm
So, CQ = BC - BQ = 38 - 27 = 11cm
⇒ CR = CQ = 11cm
In quad. SORD,
$\angle\text{SDR}=90^\circ....(\therefore\text{AD}\perp\text{CD})$
$\Rightarrow\angle\text{OSD}=\angle\text{ORD}=90^\circ$
Also, OS = OR and SD = SR
So, quad. SORD is a square.
Thus, DR = SO = 10cm
Hence, CD = DR + CR = 10 + 11 = 21cn.
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MCQ 141 Mark
In the given figure, O is the centre of a circle. AOC is its diameter, such that $\angle\text{ACB} = 50^\circ.$ If AT is the tangent to the circle at the point A, then$\angle \text{BAT} = ?$
  • A
    40°
  • ✓
    50°
  • C
    60°
  • D
    65°
Answer
Correct option: B.
50°
Construction: Join OC.Since AC is a diameter of the circle.
$\angle\text{ABC}=90^\circ$ ....(angle in a semicircle is 90º)
In $\triangle\text{ABC},$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ$ ......(Angle Sum Property)
$\Rightarrow90^\circ+50^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BAC}=40^\circ$
Since AC is a tangent to the inner circle.
$\Rightarrow\angle\text{OAT}=90^\circ$ .....(Tangent is perpendicular to the radius of a circle)
$\Rightarrow\angle\text{BAO}+\angle\text{BAT}=90^\circ$
$\Rightarrow40^\circ+\angle\text{BAT}=90^\circ$
$\Rightarrow\angle\text{BAT}=50^\circ$
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MCQ 151 Mark
PQ is a tangent to a circle with centre O at the point P. If $\triangle\text{OPQ}$ is an isosceles triangle, then $\angle\text{OQP}$ is equal to:
  • A
    30º
  • ✓
    45º
  • C
    60º
  • D
    90º
Answer
Correct option: B.
45º



Given that $\triangle\text{PQO}$ is an isosceles triangle.

Since PQ is a tangent to the circle at P.

Sunce PQ is a tangent to the circle at P.

$\angle\text{OPQ}=90^\circ$ .....(tangent is perpendicular to the radius of a circle)

In $\triangle\text{OPQ},$

OP = OQ

$\Rightarrow\angle\text{OQP}=\angle\text{POQ}$

Using Angle Sum Property,

$\angle\text{OQP}+\angle\text{POQ}+\angle\text{OPQ}=180^\circ$

$\Rightarrow\angle\text{OQP}+\angle\text{OQP}+90^\circ=180^\circ$

$\Rightarrow2\angle\text{OQP}=90^\circ$

$\Rightarrow\angle\text{OQP}=45^\circ$
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MCQ 161 Mark
Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:
Assertion (A)
Reason (R)
In the given figure, a quad. ABCD is drawn to circumscribe a given circle as shown. Then, AB + BC = AD + DC. In two concentric circles, the chord of the larger circle, which to uches the smaller circle, is bisected at the point of contact.
  • A
    Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
  • B
    Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
  • C
    Assertion (A) is true and Reason (R) is false.
  • ✓
    Assertion (A) is false and Reason (R)is true.
Answer
Correct option: D.
Assertion (A) is false and Reason (R)is true.
The Assertion (A) is false since sum of the opposite sides of a quadrilateral circumscribing a circle are equal, and not the adjacent sides.The chord of the larger circle is the tangent to the smaller circle.
We know that the perpendicular drawn from the centre to the chord
So, the Reason (R) is true.
But is not the correct explanation for the Assertion (A).
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MCQ 171 Mark
If the angle between two radii of a circle is 130°, then the angle between the tangent at the ends of the radii is:
  • A
    65º
  • B
    40º
  • ✓
    50º
  • D
    90º
Answer
Correct option: C.
50º

In quad. AOBP
$\angle\text{PAO}+\angle\text{PBO}+\angle\text{AOB}+\angle\text{APB}=360^\circ$....(Angle Sum Property)
$\Rightarrow90^\circ+90^\circ+130^\circ+\angle\text{APB}=360^\circ$....(Since radius of a circle is perpendicular to the tangent)
$\Rightarrow\angle\text{APB}=50^\circ$
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MCQ 181 Mark
In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. $\angle\text{PBO} = 30^\circ$ then $\angle\text{PTA} =?$
  • A
    60º
  • ✓
    30º
  • C
    15º
  • D
    45º
Answer
Correct option: B.
30º
In $\triangle\text{OBP},$$\text{OB} = \text{OP} $....(radii of the circle)
$\Rightarrow\angle\text{OBP}=\angle\text{OPB}=30^\circ$ ....(angles opposite equal sides are equal)
Since PT is a tangent,
$\angle\text{OPT}=90^\circ$
In $\triangle\text{BPT},$
$\angle\text{BPT}+\angle\text{PBT}+\angle\text{PTB}=180^\circ$....(Angle Sum Property)
$\Rightarrow(30^\circ+90^\circ)+30^\circ+\angle\text{PTB}=180^\circ$
$\Rightarrow\angle\text{PTB}=30^\circ$
That is, $\angle\text{PTA}=30^\circ$
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MCQ 191 Mark
$O$ is the centre of a circle of radius $5\ cm.$ At a distance of $13\ cm$ from $O$, a point $P$ is taken. From this point, two tangents $PQ$ and $PR$ are drawn to the circle. Then, the area of quadrilateral $\text{PQOR}$ is:
  • ✓
    $60\ cm^2$
  • B
    $32.5\ cm^2$
  • C
    $65\ cm^2$
  • D
    $30\ cm^2$
Answer
Correct option: A.
$60\ cm^2$
In $\triangle\text{OPQ}$ and $\triangle\text{ORP},$
$\angle\text{OQP}=\angle\text{ORP}=90^\circ ....($Since $OP$ and $RP$ are tangent to the circle$)$
$\text{OP}=\text{OP} ....($common side$)$
$\text{OQ}=\text{OR} ....($radii of the same circle$)$
$\Rightarrow\triangle\text{OPQ}\cong\triangle\text{ORP} ....(\text{RHS}$ congruence criterion$)$
So, the areas of both the triangle will be the swame.
In $\triangle\text{OPQ},$
By Pythagoras theorem,
$\text{OP}^2=\text{OQ}^2+\text{PQ}^2$
$\Rightarrow\text{PQ}^2=\text{OP}^2-\text{OQ}^2$
$\Rightarrow\text{PQ}^2=\text{13}^2-\text{5}^2$
$\Rightarrow\text{PQ}^2=144$
$\Rightarrow\text{PQ}=12\text{cm}$
$\text{ar}(\triangle\text{OPQ})=\frac{1}{2}\times\text{PQ}\times\text{OQ}$
$=\frac{1}{2}\times\text{12}\times5$
$=30\text{cm}^2$
$\text{ar}(\text{quad PQOR})=\text{ar}(\triangle\text{OPQ})+\text{ar}(\triangle\text{ORP})$
$\Rightarrow\text{ar}(\text{quad PQOR})=30\text{cm}^2+30\text{cm}^2$
$\Rightarrow\text{ar}(\text{quad PQOR})=60\text{cm}^2$
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MCQ 201 Mark
In the given figure, PA and PB are two tangents to the circle with centre O. If $\angle\text{APB}=60^\circ$ then $\angle\text{OAB}$ is:
  • A
    15º
  • ✓
    30º
  • C
    60º
  • D
    90º
Answer
Correct option: B.
30º
We know that tangent from an external point to a circle are equal.So,
$\text{PA}=\text{PB}$
$\Rightarrow\angle\text{PAB}+\angle\text{PBA}$ ....(angles opposite equal sides are equal)
Now in $\triangle\text{PAB},$
$\angle\text{PAB}+\angle\text{PBA}+\angle\text{APB}=180^\circ$ ...(angles Sum Property)
$\Rightarrow\angle\text{PAB}+\angle\text{PAB}+60^\circ=180^\circ$
$\Rightarrow2\angle\text{PAB}+60^\circ=180^\circ$
$\Rightarrow2\angle\text{PAB}=120^\circ$
$\Rightarrow\angle\text{PAB}=60^\circ$
Since AP is a tangent to the circle,
$\angle\text{OAP}=90^\circ$
$\Rightarrow\angle\text{OAB}+\angle\text{PAB}=90^\circ$
$\Rightarrow\angle\text{OAB}+60^\circ=90^\circ$
$\Rightarrow\angle\text{OAB}=30^\circ$
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MCQ 211 Mark
In the given figure, point $P$ is $26\ cm$ away from the centre $O$ of a circle and the length $PT$ of the tangent drawn from $P$ to the circle is $24\ cm.$ Then the radius ofthe circle is:
  • ✓
    $10\ cm$
  • B
    $12\ cm$
  • C
    $13\ cm$
  • D
    $15\ cm$
Answer
Correct option: A.
$10\ cm$
Construction: Join $OT.$
$PT = 24\ cm$
$OP = 26\ cm$
Sunce $PT$ is a tangent to the circle at $T.$
$\angle\text{PTO}=90^\circ .....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{PTO},$
By Pythagoras theorem,
$OP^2=PT^2+OT^2$
$\Rightarrow OT^2=OP^2-PT^2$
$\Rightarrow OT^2=26^2-24^2$
$\Rightarrow OT^2=676-576$
$\Rightarrow O T^2=100$
$\Rightarrow OT^2=10 \ cm$
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MCQ 221 Mark
In the given figure, AD and AE are the tangents to a circle with centre O and BC touches the circle at F. If AE = 5cm, then perimeter of $\triangle\text{ABC}$ is:
  • A
    15cm
  • ✓
    10cm
  • C
    22.5cm
  • D
    20cm
Answer
Correct option: B.
10cm
We know that tangent from an external point to the circle equal.So,
AE = AD = 5cm
BF = BE
CF = CD
Perimeter of $\triangle\text{ABC}$
= AB + BC + AC
= AB + (BE + DC) + AC
= AB + (BE + DC) + AC
= (AB + BE) + (AC + DC)
= AE + AD
= 5 + 5
= 10cm
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MCQ 231 Mark
Which of the following statements is not true?
  • A
    A line which intersects a circle at two points, is called a secant of the circle.
  • ✓
    A line intersecting a circle at one point only, is called a tangent to the circle.
  • C
    The point at which a line touches the circle, is called the point of contact.
  • D
    A tangent to the circle can be drawn from a point inside the circle.
Answer
Correct option: B.
A line intersecting a circle at one point only, is called a tangent to the circle.
Options (a), (b) and (c) are all true.However, Option (d) is false since it is not possible to draw a tangent from a point inside a circle.
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MCQ 241 Mark
In the given figure, $AB$ and $AC$ are tangents to a circle with centre $O$ and radius $8\ cm$. If $OA = 17\ cm$, then the length of $AC ($ in $cm)$ is:
  • A
    $9$
  • ✓
    $15$
  • C
    $\sqrt{353}$
  • D
    $25$
Answer
Correct option: B.
$15$

Construction: Join $OC.$
Since $AC$ is a tangent to the circle.
$\angle\text{OCA}=90^\circ ....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{OCA},$
By Pythagoras theorem,
$OA^2 = OC^2 + AC^2$
$\Rightarrow 17^2 = 8^2 + AC^2$
$\Rightarrow AC^2 = 289 - 64$
$\Rightarrow AC^2 = 225$
$\Rightarrow AC = 15\ cm$
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MCQ 251 Mark
In a circle of radius $7\ cm,$ tangent $PT$ is drawn from a point $P,$ such that $PT = 24\ cm.$ If $O$ is the centre of the circle, then $OP =?$
  • A
    $30\ cm$
  • B
    $28\ cm$
  • ✓
    $25\ cm$
  • D
    $18\ cm$
Answer
Correct option: C.
$25\ cm$
$PT = 24\ cm, OT = 7\ cm.$
Since $PT$ is a tangent to the circle at $T.$
$\angle\text{PTO}=90^\circ ....($tangent is perpendicular to the radius of a circle$)$
In $\triangle\text{PTO},$
By using Pythagoras theorem,
$OP^2 = PT^2 + OT^2$
$\Rightarrow OP^2 = 24^2+ 7^2$
$\Rightarrow OP^2 = 576 + 49$
$\Rightarrow OP^2 =625$
$\Rightarrow OP = 25\ cm.$
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MCQ 261 Mark
In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of $\angle\text{ACB}$ is:
  • A
    45º
  • B
    60º
  • ✓
    90º
  • D
    120º
Answer
Correct option: C.
90º

Draw a tangent to the circle at point C meet AB at P.
Then,
PA = PC
⇒ $\angle\text{PAC}=\angle\text{PCA}$
And PB = PC
$\Rightarrow\angle\text{PBC}=\angle\text{PCB}$
$\therefore\angle\text{PAC}+\angle\text{PBC}=\angle\text{PCA}+\angle\text{PCB}=\angle\text{ACB}$
$\Rightarrow\angle\text{PAC}+\angle\text{PBC}+\angle\text{ACB}=2\angle\text{ACB}$
$\Rightarrow180^\circ=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=90^\circ$
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MCQ 271 Mark
In the given figure, O is the centre of the circle, PQ is a chord and PT is the tangent at P. If $ \angle\text{POQ} = 70^\circ$then $\angle\text{TPQ} $ is equal to:
  • ✓
    35º
  • B
    45º
  • C
    55º
  • D
    70º
Answer
Correct option: A.
35º
In $\triangle\text{OPQ},$$\text{OP}=\text{OQ}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OQP}=\angle\text{OPQ}$ ....(angles opposite equal sides are equal)
$\triangle\text{OPQ},$
$\angle\text{OQP}+\angle\text{OPQ}+\angle\text{POQ}=180^\circ$ ......(Angle Sum Property)
$\Rightarrow\angle\text{OPQ}+\angle\text{OPQ}+70^\circ=180^\circ$
$\Rightarrow2\angle\text{OPQ}=110^\circ$
$\Rightarrow\angle\text{OPQ}=55^\circ$
Since PT is a tangent to the inner circle.
$\Rightarrow\angle\text{OPT}=90^\circ$ .....(Tangent is perpendicular to the radius of a circle)
$\Rightarrow\angle\text{OPQ}+\angle\text{TPQ}=90^\circ$
$\Rightarrow55^\circ+\angle\text{TPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}=35^\circ$
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MCQ 281 Mark
The number of tangents that can be drawn from an external point to a circle is:
  • A
    1
  • ✓
    2
  • C
    3
  • D
    4
Answer
Correct option: B.
2

We can draw only two tangents from an external point to a circle.
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MCQ 291 Mark
In the given figure, if $\angle\text{AOD}=135^\circ$ then $\angle\text{BOC}$ is equal to:
  • A
    25º
  • ✓
    45º
  • C
    52.5º
  • D
    62.5º
Answer
Correct option: B.
45º
We know that sum of the angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180º.$\Rightarrow\angle\text{AOD}+\angle\text{BOC}=180^\circ$
$\Rightarrow135^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=45^\circ$
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MCQ 301 Mark
In the given figure, the length of BC is:
  • A
    7cm
  • ✓
    10cm
  • C
    14cm
  • D
    15cm
Answer
Correct option: B.
10cm
We know that tangent from an external point to a circle are equal.
So,
AF = AE = 4cm
⇒ EC = AC - AE = 11 - 4 = 7cm
Now,
CD = CE = 7cm
and BF = BD = 3cm
BD = BD + CD
⇒ BD = 3 + 7
⇒ BD = 10cm
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MCQ 311 Mark
In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If $\angle\text{PAB} = 67^\circ$then the measure of $\angle\text{AQB} $ is:
  • A
    73º
  • B
    64º
  • C
    53º
  • ✓
    44º
Answer
Correct option: D.
44º
Since $\angle\text{BAC}$ is inscribed in a semicircle, $\angle\text{BAC}=90^\circ.$
Since PAQ is a straight line,
$\angle\text{PAB}+\angle\text{BAC}+\angle\text{CAQ}=180º$
$\Rightarrow67^\circ+90^\circ+\angle\text{CAQ}=180^\circ$
$\Rightarrow\angle\text{CAQ}=23^\circ$
We know that angles that subtend the same arc are equal.
$\Rightarrow\angle\text{CBA}=\angle\text{CAQ}=23^\circ$
In $\triangle\text{BAQ},$
$\angle\text{BAQ}+\angle\text{QBA}+\angle\text{AQB}=180º$....(Angle Sum Property)
$\Rightarrow(90^\circ+23^\circ)+23^\circ+\angle\text{AQB}=180^\circ$
$\Rightarrow\angle\text{AQB}=44^\circ$
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MCQ 321 Mark
Which of the following statment is not true?
  • A
    If a point P lies inside a circle, no tangent can be drawn to the circle passing through P.
  • B
    If a point P lies on a circle, then one and only one tangent can be drawn to the circle at P.
  • C
    If a point P lies outside a circle, then only two tangents can be drawn to the circle from P.
  • ✓
    A circle can have more than two parallel tangents parallel to a given line.
Answer
Correct option: D.
A circle can have more than two parallel tangents parallel to a given line.
Options (a), (b) and (c) are all true.However, Option (d) is false since we can draw only parallel tamngent on either side of the diameter, which would be parallel to a given line.
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MCQ 331 Mark
To draw a pair of tangents to a circle, which are inclined to each other at angle of 45º, we have to draw the tangents at the end points of those two radii, the angle between which is:
  • A
    105º
  • ✓
    135º
  • C
    140º
  • D
    145º
Answer
Correct option: B.
135º


Since AB and AC are the tangent to the circle.
$\angle\text{OBA}=\angle\text{OCA}=90^\circ$ (tangent is perpendicular to the radius of a circle)
In ACOB,
$\angle\text{OBA}+\angle\text{BAC}+\angle\text{OCA}+\angle\text{BOC}=360^\circ$
$\Rightarrow90^\circ+45^\circ+45^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=135^\circ$
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MCQ 341 Mark
If two tangents inclined at an angle of 60° are drawn to a circle of radius 3cm, then the length of each tangent is:
  • A
    $3\text{cm}$
  • B
    $\frac{3\sqrt{3}}{2}\text{cm}$
  • ✓
    $3\sqrt{3}\text{cm}$
  • D
    $6\text{cm}$
Answer
Correct option: C.
$3\sqrt{3}\text{cm}$



In $\triangle\text{BAO}$ and $\triangle\text{CAO},$

$\angle\text{OBA}=\angle\text{OCA}=90^\circ$ ....(Since AB and AC are tangent to the circle)

$\text{OA}=\text{OA}$ ....(common side)

$\text{OB}=\text{OC}$ ....(radii of the same circle)

$\Rightarrow\triangle\text{BAO}\cong\triangle\text{CAO}$ ....(RHS congruence criterion)

$\angle\text{OAB}=\angle\text{OAC}$ ....(cpct)

$\Rightarrow\angle\text{OAB}=\frac{1}{2}\angle\text{BAC}=30^\circ$

So, $\triangle\text{BAO}$ is a 30-60-90 triangle.

side opposite $30^\circ=\frac{1}{2}$ hypotenuse

$\Rightarrow\text{OB}=\frac{1}{2}$ hypotenuse

⇒ hypotene = 2OB = 2(3) = 6cm

side opposite $60^\circ=\frac{\sqrt{3}}{2}$ hypotenuse

$\Rightarrow\text{AB}=\frac{\sqrt{3}}{2}(6)=3\sqrt{3}\text{cm}$

$\text{AB}=\text{AC}=3\sqrt{3}\text{cm}$ ....(Since tangents from an external point to the circle are equal)

Hence, the length of each tangent is $3\sqrt{3}\text{cm}.$
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MCQ 351 Mark
In the given figure, PQ and PR are tangents to a circle with centre A. If $\angle\text{QPA}=27^\circ$ then $\angle\text{QAR}$ equals:
  • A
    63º
  • B
    117º
  • ✓
    126º
  • D
    153º
Answer
Correct option: C.
126º
In $\triangle\text{PAQ}$ and $\triangle\text{PAR},$
$\angle\text{PQA}=\angle\text{PRA}$ ....(Since PQ and PR are tangent to the circle)
$\text{AP}=\text{AP}$ ....(common side)
$\text{AQ}=\text{AR}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{PAQ}\cong\triangle\text{PAR}$ ....(RHS congruence criterion)
$\angle\text{OAP}=\angle\text{RAP}$ ....(cpct)
In $\triangle\text{PAQ},$
$\angle\text{QAP}+\angle\text{PQA}+\angle\text{APQ}=180^\circ$ ....(Angle Sum property)
$\Rightarrow\angle\text{QAP}+90^\circ+27^\circ=180^\circ$.....$(\angle\text{PQA}=90^\circ,$ since radius is perpendicular to the tangent $)$
$\Rightarrow\angle\text{QAP}=63^\circ$
So, $\angle\text{QAR}=\angle\text{QAP}+\angle\text{RAP}$
$\Rightarrow\angle\text{QAR}=63^\circ+63^\circ$
$\Rightarrow\angle\text{QAR}=126^\circ$
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MCQ 361 Mark
In the given figure,$ \triangle\text{ABC}$ is right-angled at B, such that BC = 6cm and AB = 8cm. A circle with centre O has been inscribed in the triangle. $\text{OP} \perp \text{AB}, \text{OQ} \perp \text{BC} $ and $\text{OR} \perp \text{AC}.$ If OP = OQ = OR = x cm, then x =?
  • ✓
    2cm
  • B
    2.5cm
  • C
    3cm
  • D
    3.5cm
Answer
Correct option: A.
2cm
In right $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ ....(By Pythagoras theorem)
$\Rightarrow\text{AC}^2=8^2+6^2$
$\Rightarrow\text{AC}^2=64+36$
$\Rightarrow\text{AC}=100$
$\Rightarrow\text{AC}=10\text{cm}$
We know that tangent from an external poit to the circle are equal.
$\Rightarrow\text{CR}=\text{CQ}=\text{BC}-\text{BQ}=(6-\text{x})\text{cm}$
$\Rightarrow\text{AR}=\text{AP}=\text{AB}-\text{BP}=(8-\text{x})\text{cm}$
$\text{AC}=(\text{AR}+\text{CR})=(8-\text{x})+(6-\text{x})=(14-2\text{x})\text{cm}$
$\Rightarrow14-2\text{x}=10$
$\Rightarrow2\text{x}=4$
$\Rightarrow\text{x}=2$
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MCQ 371 Mark
In the given figure, DE and DF are tangents from an external point D to a circle with centre A. If DE = 5cm. and $\text{DE}\perp\text{DF}$ then the radius of the circle is:
  • A
    3cm
  • B
    4cm
  • ✓
    5cm
  • D
    6cm
Answer
Correct option: C.
5cm

Construction: Join AE and AF.
Since DE and DF are tangent to the circle,
$\angle\text{AED}=\angle\text{AFD}=\angle\text{EDF}=90^\circ$
Also, AE = AF .....(radii of the same circle)\
and ED = EF ....(Since tangent drawn from an external point to the circle are equal)
So, quadrilateral AEDF is a square.
Thus, AE = DF = 5cm
Hence, the length of the radius is 5cm.
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MCQ 381 Mark
In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then,$ \angle\text{POQ} = ?$
  • A
    130º
  • ✓
    100º
  • C
    90º
  • D
    75º
Answer
Correct option: B.
100º
Since PT is the tangent to the circle.$\angle\text{OPT}=90^\circ$
$\Rightarrow\angle\text{TPQ}+\angle\text{OPQ}=90^\circ$
$\Rightarrow50^\circ+\angle\text{OPQ}=90^\circ$
$\Rightarrow\angle\text{OPQ}=40^\circ$
In $\triangle\text{OPQ},$
$\text{OP}=\text{OQ}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OPQ}=\angle\text{OQP}=40^\circ$....(angles opposite equal sides are equal)
Now,
$\angle\text{OPQ}+\angle\text{OQP}+\angle\text{POQ}=180^\circ$....(Angle Sum Property)
$\Rightarrow40^\circ+40^\circ+\angle\text{POQ}=180^\circ$
$\Rightarrow\angle\text{POQ}=100^\circ$
View full question & answer→
MCQ 391 Mark
Which of the following pairs of lines in a circle cannot be parallel?
  • A
    Two chords
  • B
    A chord and a tangent
  • C
    Two tangents
  • ✓
    Two diameters
Answer
Correct option: D.
Two diameters
Two diameter of the circle always passes through the centre. This means all the diameters of a given circle will intersect at the centre, and hence they cannot be parallel.
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MCQ 401 Mark
In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6cm such that the segments QT and TR into which QR is divided by the point of contact T are of lengths 12cm and 9cm respectively. If the area of $\triangle\text{PQR} = 189 \text{cm}^2 $ then the length of side PQ is:
  • A
    17.5cm
  • B
    20cm
  • ✓
    22.5cm
  • D
    25cm
Answer
Correct option: C.
22.5cm



We know that tangent from an external point to the circle are equal.

So,

QT = QN = 12cm

TR = RM = 9cm

Now,

$\text{ar}(\triangle\text{PQR})=\frac{1}{2}(\text{Perimeter of }\triangle\text{PQR})\times\text{r}$

$\Rightarrow\text{ar}(\triangle\text{PQR})=\frac{1}{2}(12+12+9+9+\text{x}+\text{x})\times\text{r}$

$\Rightarrow\text{ar}(\triangle\text{PQR})=\frac{1}{2}(42+2\text{x})\times6$

$\Rightarrow189=3(42+2\text{x})$

$\Rightarrow63=42+2\text{x}$

$\Rightarrow2\text{x}=21$

$\Rightarrow\text{x}=10.5$

So, PQ = 12 + 10.5 = 22.5cm
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MCQ 411 Mark
If tangents PA and PB from a point P to a circle with centre O are drawn, so that $\angle\text{APB} = 80^\circ, $ then $\angle\text{POA} = ?$
  • A
    40º
  • ✓
    50º
  • C
    80º
  • D
    60º
Answer
Correct option: B.
50º
In $\triangle\text{PAO}$ and $\triangle\text{PBO},$$\angle\text{PAO}+\angle\text{PBO}=90^\circ$....(Since PQ and PR are tangent to the circle)
$\text{OP}=\text{OP}$ ....(Common side)
$\text{AO}=\text{OB}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{PAO}\cong\triangle\text{PBO}$ ....(RHS congruence criterion)
$\angle\text{POA}=\angle\text{BOP}$ ....(cpct)
In quad. AOBP
$\angle\text{PAO}+\angle\text{PBO}+\angle\text{AOB}+\angle\text{APB}=360^\circ$
$\Rightarrow90^\circ+90^\circ+\angle\text{AOB}+80^\circ=360^\circ$
$\Rightarrow\angle\text{AOB}=100^\circ$
So, $\angle\text{POA}=50^\circ$
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MCQ 421 Mark
In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord such that $\angle\text{QPT}=50^\circ$ then $\angle\text{POQ}=?$
  • ✓
    100º
  • B
    90º
  • C
    80º
  • D
    75º
Answer
Correct option: A.
100º
Since PT is the tangent to the circle,$\angle\text{TPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}+\angle\text{OPQ}=90^\circ$
$\Rightarrow50^\circ+\angle\text{OPQ}=90^\circ$
$\Rightarrow\angle\text{OPQ}=40^\circ$
In $\triangle\text{OPQ},$
$\text{OP}=\text{OQ}$ ....(radii of the same circle)
$\Rightarrow\angle\text{OPQ}=\angle\text{OQP}=40^\circ$ ....(angles opposite equal sides are equal)
Now,
$\angle\text{OPQ}+\angle\text{OQP}+\angle\text{POQ}=180^\circ$ ....(Angle Sum Property)
$\Rightarrow40^\circ+40^\circ+\angle\text{POQ}=180^\circ$
$\Rightarrow\angle\text{POQ}=100^\circ$
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MCQ 431 Mark
In the given figure, a circle touches the side DF of $\triangle\text{EDF}$ at H and touches ED and EF produced at K and M respectively. If EK = 9cm then the perimeter of $\triangle\text{EDF}$ is:
  • A
    9cm
  • B
    12cm
  • C
    13.5cm
  • ✓
    18cm
Answer
Correct option: D.
18cm
We know that tangents from an external point to the circle are equal.
So,
EK = EM = 9cm
DK = DH
FH = FM
perimeter of $\triangle\text{EDF}$
= ED + EF + DF
= ED + EF + DH + HF
= (ED + DH) + (EF + HF)
= (ED + DK) + (EF + FM)
= EK + EM
= 9 + 9
= 18cm.
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MCQ 441 Mark
In the given figure, O is the centre of two concentric circles of radii 6cm and 10cm. AB is a chord of outer circle which touches the inner circle. The length of AB is:
  • A
    8cm
  • B
    14cm
  • ✓
    16cm
  • D
    $\sqrt{136}\text{cm}$
Answer
Correct option: C.
16cm
Since AB is a tangent to the circle.$\angle\text{OPA}=90^\circ$ ....(tangent is perpendicular to the radius of a circle)
AB is a chord of the outer circle
We know that of the perpendicular drawn from the centre to a chord of a circle of a circle, bisects the chord.
In $\triangle\text{OPA},$
$AO^2 = OP^2 + AP^2$
$\Rightarrow 10^2 = 6^2 + AP^2$
$\Rightarrow AP^2 = 10^2 - 6^2$
$\Rightarrow PA^2 = 64$
$\Rightarrow PA = 8cm$
$AB = 2AP = 2(8) = 16cm.$
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MCQ 451 Mark
In the given figure, AT is a tangent to the circle with centre O, such that OT = 4cm and $\angle\text{OTA} = 30^\circ. $Then, AT =?
  • A
    $4\text{cm}$
  • B
    $2\text{cm}$
  • ✓
    $2\sqrt{3}\text{cm}$
  • D
    $4\sqrt{3}\text{cm}$
Answer
Correct option: C.
$2\sqrt{3}\text{cm}$
Since $\angle\text{OAT}=90^\circ$ and $\angle\text{OTQ}=30^\circ$
Clearly, $\angle\text{AOT}=60^\circ$
So, $\triangle\text{AOT}$ is a 30º - 60º - 90º triangle.
Side opposite $60^\circ=\frac{\sqrt{3}}{2}$ hypotenuse
$\Rightarrow\text{AT}=\frac{\sqrt{3}}{2}(\text{OT})$
$\Rightarrow\text{AT}=\frac{\sqrt{3}}{2}(4)$
$\Rightarrow\text{AT}=2\sqrt{3}\text{cm}$
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MCQ 461 Mark
Which of the following statements is not true?
  • A
    A tangent to a circle intersects the circle exactly at one point.
  • B
    The point common to a circle and its tangent is called the point of contact.
  • C
    The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
  • ✓
    A straight line can meet a circle at one point only.
Answer
Correct option: D.
A straight line can meet a circle at one point only.
Options (a), (b) and (c) are all true.
However, Option (d) is false since a straight line can meet a circle at two points even as shown below.
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MCQ 471 Mark
Quadrilateral ABCD is circumscribed to a circle. If AB = 6cm, BC = 7cm and CD = 4cm, then the length of AD is:
  • ✓
    3cm
  • B
    4cm
  • C
    6cm
  • D
    7cm
Answer
Correct option: A.
3cm



Using the property, tangent from an external point to the circle are equal.

We can say, AB + CD = AD + BC

⇒ AD = AB + CD - BC

⇒ AD = 6 + 4 - 7

⇒ AD = 3cm
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MCQ 481 Mark
The chord of a circle of radius $10\ cm$ subtends a right angle at its centre. The length of the chord $($in $cm)$ is:
  • A
    $\frac{5}{\sqrt{2}}\text{cm}$
  • B
    $5\sqrt{2}\text{cm}$
  • ✓
    $10\sqrt{2}\text{cm}$
  • D
    $10\sqrt{3}\text{cm}$
Answer
Correct option: C.
$10\sqrt{2}\text{cm}$

In $\triangle\text{POQ},$
By using Pythagoras theorem,
$PQ^2 = PO^2 + OQ^2$
$\Rightarrow PQ^2= 10^2 + 10^2$
$\Rightarrow PQ^2= 100 + 100$
$\Rightarrow PQ^2= 200$
$\Rightarrow PQ^2 = 200$
$\Rightarrow\text{PQ}=10\sqrt{2}\text{cm}$
So, the length of the chord is $10\sqrt{2}\text{cm}$
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MCQ 491 Mark
In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR, such that$ \angle\text{BQR} = 70^\circ.$ Then, $\angle\text{AQB}=?$
  • A
    20º
  • B
    35º
  • ✓
    40º
  • D
    45º
Answer
Correct option: C.
40º
Since AB || PQ
$\angle\text{BQR}=\angle\text{ABQ}=70^\circ$ ....(alternate angles)
and $\angle\text{PQA}=\angle\text{BAQ}=70^\circ$....(alternate angles)
In $\triangle\text{ABQ},$
$\angle\text{ABQ}+\angle\text{BAQ}+\angle\text{AQB}=180^\circ$....(angle Sum Property)
$\Rightarrow70^\circ+70^\circ+\angle\text{AQB}=180^\circ$
$\Rightarrow\angle\text{AQB}=40^\circ$
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MCQ 501 Mark
In the given figure, three circles with centres A, B, C, respectively, touch each other externally. If AB = 5cm, BC = 7cm and CA = 6cm, the radius of the circle with centre A is:
  • A
    1.5cm
  • ✓
    2cm
  • C
    2.5cm
  • D
    3cm
Answer
Correct option: B.
2cm
Let the radii of the circle with centres A, B and C be x, y, and z respectively.
We know that radii of the same circle are equal.
x + y = 5
y + z = 7
z + x = 6
Adding the three equation, we get
2(x + y + z) = 18
⇒ x + y + z = 9
⇒ x = 2
So, the radius of the circle with centre A is 2cm.
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MCQ 511 Mark
In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4cm. If $\text{PA}\perp \text{PB}$ then the length of each tangent.
  • A
    3cm
  • ✓
    4cm
  • C
    5cm
  • D
    6cm
Answer
Correct option: B.
4cm
Construction: Join CA and CB.Since AP and PB are tangent to the circle,
$\angle\text{CAP}=\angle\text{CBA}=90^\circ$
Given that $\angle\text{APB}=90^\circ$
We know that tangents drawn from an external point to the circle are equal.
$\Rightarrow\text{AP}=\text{PB}$
Also, $\text{CA}=\text{CB}$ ....(radii of the same circle)
So, quadrilateral APBC is a square.
Thus, AP = PB = CA = CB = 4cm.
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MCQ 521 Mark
If a chord $AB$ subtends an angle of $60^\circ$ at the centre of a circle, then the angle between the tangents to the circle drawn from $A$ and $B$ is:
  • A
    $30^\circ $
  • B
    $60^\circ $
  • C
    $90^\circ $
  • ✓
    $120^\circ $
Answer
Correct option: D.
$120^\circ $

Since $AD$ and $DB$ are the tangent to the circle.
$\angle\text{OAD}=\angle\text{OBD}=90^\circ....($tangent is perpendicular to the radius of a circle$)$
In $\text{ABOD}$
$\angle\text{OAD}+\angle\text{ADB}+\angle\text{OBD}+\angle\text{AOB}=360^\circ$
$\Rightarrow90^\circ+\angle\text{ADB}+90^\circ+60^\circ=360^\circ$
$\Rightarrow\angle\text{ADB}=120^\circ$
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MCQ 531 Mark
In a right triangle ABC, right angled at B, BC = 12cm and AB = 5cm. The radius of the circle inscribed in the triangle is:
  • A
    1cm
  • ✓
    2cm
  • C
    3cm
  • D
    4cm
Answer
Correct option: B.
2cm
In right $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ ....(By Pythagoras theorem)
$\Rightarrow\text{AC}^2=5^2+12^2$
$\Rightarrow\text{AC}^2=169$
$\Rightarrow\text{AC}=13\text{cm}$
We know that,
$\text{ar}(\triangle\text{ABC})=\frac{1}{2}\times(\text{perimeter of }\triangle\text{ABC})\times\text{r}$
$\Rightarrow\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times(\text{perimeter of }\triangle\text{ABC})\times\text{r}$
$\Rightarrow\frac{1}{2}\times12\times5=\frac{1}{2}\times(5+12)\times\text{r}$
$\Rightarrow\frac12\times5=30\times\text{r}$
$\Rightarrow\text{r}=2\text{cm}$
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MCQ 541 Mark
In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6cm, BP = 5cm, CQ = 3cm and DR = 4cm, then the perimeter of quadrilateral ABCD is:
  • A
    18cm
  • B
    27cm
  • ✓
    36cm
  • D
    32cm
Answer
Correct option: C.
36cm
We know that tangent from an external point to the circle are equal.RC = QC = 3cm
PB = BQ = 5cm
AP = AS = 6cm
SD = DR = 4cm
Perimeter of quad. ABCD
= AB + BC + CD + AD
= (AP + PB) + (BQ + CQ) + (CR + DR) + (AS + SD)
= (6 + 5) + (5 + 3) + (3 + 4) + (6 + 4)
= 36cm
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MCQ 551 Mark
In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If $\angle\text{APQ}=58^\circ$ then the measure of $\angle\text{PQB}$ is:
  • ✓
    32º
  • B
    58º
  • C
    122º
  • D
    132º
Answer
Correct option: A.
32º
$\angle\text{APQ}=58^\circ$
$\angle\text{QPR}=90^\circ$ ....(angle inscribed a semicircle)
Since APB is a straight line,
$\angle\text{APQ}+\angle\text{QPR}+\angle\text{RPB}=180^\circ$
$\Rightarrow58^\circ+90^\circ+\angle\text{RPB}=180^\circ$
$\Rightarrow\angle\text{RPB}=32^\circ$
We know that angle that subtend the same arc are equal.
So, $\angle\text{RPB}=\angle\text{RPB}=32^\circ$
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MCQ 561 Mark
In the given figure, PA and PB are tangents to the given circle, such that PA = 5cm and $\angle\text{APB} = 60^\circ.$ The length of chord AB is:
  • A
    $5\sqrt{2}\text{cm}$
  • ✓
    $5\text{cm}$
  • C
    $5\sqrt{3}\text{cm}$
  • D
    $7.5\text{cm}$
Answer
Correct option: B.
$5\text{cm}$
We know that tangents from an external point to the circle are equal.$\text{PA}=\text{PB}$
$\Rightarrow\angle\text{PBA}=\angle\text{PAB}=\text{x}^\circ$
In $\triangle\text{PAB},$
$\angle\text{PBA}+\angle\text{PAB}+\angle\text{APB}=180^\circ$....(Angle Sum Property)
$\Rightarrow\angle\text{PAB}+\angle\text{PAB}+60^\circ=180^\circ$
$\Rightarrow2\angle\text{PAB}=120^\circ$
$\Rightarrow\angle\text{PAB}=60^\circ=\angle\text{PBA}$
So, $\triangle\text{PAB}$ is an equilateral triangle. thus, AB = PA = 5cm.
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MCQ 571 Mark
If PA and PB are two tangents to a circle with centre O, such that $\angle\text{AOB} = 110^\circ,$ find $ \angle\text{APB}$ is equal to
  • A
    55º
  • B
    60º
  • ✓
    70º
  • D
    90º
Answer
Correct option: C.
70º
Since PA and PB are the tangent to the circle.
$\angle\text{OAP}=\angle\text{OBP}=90^\circ$ ...(tangent is perpendicular to the radius of a circle)
In AOBP,
$\angle\text{OAP}+\angle\text{APB}+\angle\text{OBP}+\angle\text{AOB}=360^\circ$
$\Rightarrow90^\circ+\angle\text{APB}+90^\circ+110^\circ=360^\circ$
$\Rightarrow\angle\text{APB}=70^\circ$
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MCQ 581 Mark
The length of the tangent from an external point P to a circle of radius 5cm is 10cm. The distance of the point from the centre of the circle is:
  • A
    $8\text{cm}$
  • B
    $\sqrt{104}\text{cm}$
  • C
    $12\text{cm}$
  • ✓
    $\sqrt{125}\text{cm}$
Answer
Correct option: D.
$\sqrt{125}\text{cm}$



In $\triangle\text{PTO}$

By Pythagoras theorem,

$\text{OP}^2=\text{PT}^2+\text{OT}^2$

$\Rightarrow\text{OP}^2=10^2+5^2$

$\Rightarrow\text{OP}^2=100+25$

$\Rightarrow\text{OP}=\sqrt{125}\text{cm}$

Hence, the distance of the point from the centre of the circle is $\sqrt{125}\text{cm}.$
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MCQ 591 Mark
If PA and PB are two tangent to a circle with centre O such that $\angle\text{APB}=80^\circ.$ Then, $\angle\text{AOP}=?$
  • A
    40º
  • ✓
    50º
  • C
    60º
  • D
    70º
Answer
Correct option: B.
50º
Construction: Join CA and CB.
In $\triangle\text{PAO}$ and $\triangle\text{PBO},$
$\angle\text{OAP}=\angle\text{OBP}=90^\circ$....(Since AP and PB are tangent to the circle)
$\text{OP}=\text{OP}$ ....(common side)
$\text{AO}=\text{BO}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{PAO}\cong\triangle\text{PBO}$ ....(RHS congruence criterion)
$\angle\text{OPA}=\angle\text{OPB}$ ....(cpct)
$\Rightarrow\angle\text{OPA}=\frac{1}{2}\angle\text{APB}=40^\circ$
In $\triangle\text{PAO},$
$\angle\text{OPA}+\angle\text{PAO}+\angle\text{AOP}=180^\circ$ ....(Angle Sum Property)
$\Rightarrow40^\circ+90^\circ+\angle\text{AOP}=180^\circ$
$\Rightarrow\angle\text{AOP}=50^\circ$
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