Circles — Class 10 Maths PYQs

(B) Infinite number of tangents can be drawn to a circle from a point outside the circle.

(d) $5\sqrt{3}$ cm

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(a) $2\sqrt{3}$ cm

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation for Assertion (A).

(D) Assertion (A) is false, but Reason (R) is true.

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(B) $4\sqrt{2}$ cm

(C) $9$ cm

(A) $48$ cm$^2$

(D) $6\sqrt{3}$ cm

Join $OA$ and $OC$
$OA = OC$
$\angle OAC = \angle OCA$
Also, $\angle OAB = \angle OCD$
$\Rightarrow \angle OAC + \angle OAB = \angle OCA + \angle OCD$
$\Rightarrow \angle BAC = \angle DCA$

Correct Figure
$\angle APO = 30^{\circ}$
$\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{3}{AP}$
$AP = 3\sqrt{3} \text{ cm}$

AP = $16$ cm
$\therefore OP = 16 - 10 = 6$ cm
XY $||$ CD
$\therefore \angle CPO = 90^\circ$
In right $\triangle OPC$,
$CP = \sqrt{(10)^2 - (6)^2} = 8$ cm
CD = $2 \times CP$
$= 2 \times 8 = 16$ cm

$PM = PN$
$QS = QM$
$RS = RN$ (I) (1/2 Mark)
$PM + PN = PQ + QM + PR + RN$ (II) (1/2 Mark)
$2 PM = PQ + QS + PR + RS$
$= PQ + QS + RS + PR$
$= PQ + QR + PR$ (III) (1/2 Mark)
$\therefore PM = \frac{1}{2} (PQ + QR + PR)$ (IV) (1/2 Mark)

AB = $24$ cm
OA = $15$ cm ($\frac{1}{2}$ Mark)
Join OP intersecting AB at Q. Also join OB.
$\triangle$ PAQ $\cong \triangle$ PBQ
$\therefore \angle$ PQA = $\angle$ PQB = $90^{\circ}$ and AQ = QB = $12$ cm (1 Mark)
In right $\triangle$ OQA, OQ = $9$ cm ($\frac{1}{2}$ Mark)
Now, $\triangle$ OAP $\sim \triangle$ OQA ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{OA}{OQ} = \frac{AP}{QA}$ ($\frac{1}{2}$ Mark)
$\Rightarrow AP = \frac{OA \times QA}{OQ} = \frac{15 \times 12}{9} = 20$ cm

(D) Assertion (A) is not true but Reason (R) is true.

(A) $96^\circ$

(a) $60^\circ$ (1 Mark)

(A) $25^\circ$

OA = OP
$\therefore$ In $\triangle OAP$, $\angle OPA = \angle OAP$ ... (i)
$\angle OPA + \angle APR = 90^\circ$
$\Rightarrow \angle OAP + \angle APR = 90^\circ$ Using (i)
$\Rightarrow \angle QAP + \angle APR = 90^\circ$

(C) two points

(B) $24$ cm

Perimeter of $\triangle ABC = AB + BC + CA$
$= AB + BP + CP + CA$
$= AB + BQ + CR + CA$
$[BP = BQ \; ; \; CP = CR]$
$= AQ + AR$
$= AQ + AQ$
$[AQ = AR]$
$= 2 AQ$
$\therefore AQ = \frac{1}{2}$ (Perimeter of $\triangle ABC$)

$\angle AOB = 90^\circ$
$\therefore AB^2 = 7^2 +7^2$
$\Rightarrow AB = 7\sqrt{2}$ cm
$\Rightarrow AC = \frac{7\sqrt{2}}{2}$ cm
Now In $\Delta AOC$,
$OC^2 = 7^2 - (\frac{7\sqrt{2}}{2})^2$
$\therefore OC = \frac{7\sqrt{2}}{2}$ cm

AC = $\sqrt{(15)^2 + (8)^2} = 17$ cm ($\frac{1}{2}$ Mark)
Let '$r$' be the radius of the circle.
Since, radius is perpendicular to the tangent through the point of contact.
$\therefore$ OP is perpendicular to AB and OQ is perpendicular to BC.
Thus, OPBQ is a square.
$\Rightarrow$ OP = PB = BQ = OQ = $r$ ($\frac{1}{2}$ Mark)
Thus, AR = AP = $8 - r$
and CR = CQ = $15 - r$ } (1 Mark)
Now, AC = AR + CR ($\frac{1}{2}$ Mark)
$r = 3$ cm ($\frac{1}{2}$ Mark)

Join OR.
△ AOQ ≅ △ ROQ ⇒ ∠ AOQ = ∠ ROQ$ ----- (i) $△ BOP ≅ △ ROP ⇒ ∠ BOP = ∠ ROP$ ----- (ii) Since $∠ AOR + ∠ ROB = 180^°$ $⇒ 2∠ QOR + 2∠ ROP = 180^°$ $⇒ ∠ QOR + ∠ ROP = ∠ POQ = 90^°

Correct Fig.
$\triangle$ACP $\cong \triangle$BCP
AC=BC
$\angle$ACP = $\angle$BCP
$\angle$ACP + $\angle$BCP = $180^{\circ}$
$\angle$ACP = $90^{\circ}$
$\Rightarrow$ OP $\perp$ AB
Hence OP is perpendicular bisector of chord AB.

$TP = TQ$
$\Rightarrow \angle TPQ = \angle TQP$
Let $\angle PTQ$ be $\theta$
$\Rightarrow \angle TPQ = \angle TQP = \frac{180^{\circ} - \theta}{2} = 90^{\circ} - \frac{\theta}{2}$
Now $\angle OPT = 90^{\circ}$
$\Rightarrow \angle OPQ = 90^{\circ} - (90^{\circ} - \frac{\theta}{2}) = \frac{\theta}{2}$
$\angle PTQ = 2 \angle OPQ$

Join $OA, OB, OC$ and draw $OE \perp AC$ and $OF \perp AB$.
$BF = 10$ cm, $CE = 8$ cm, Let $AF = AE = x$
$ar \triangle ABC = ar \triangle BOC + ar \triangle COA + ar \triangle AOB$
$90 = \frac{1}{2} \cdot 4 (BC + CA + AB)$
$90 = 2(18 + 8 + x + 10 + x)$
$90 = 4(18 + x)$
$x = 4.5$
$AB = 14.5$ cm and $AC = 12.5$ cm

Area ($\triangle$ ABC) = Area ($\triangle$ BOC) + Area ($\triangle$ AOC) + Area ($\triangle$ AOB)
$= \frac{1}{2} \times 4 \times 14 + \frac{1}{2} \times 4 \times (6 + x) + \frac{1}{2} \times 4 \times (8 + x)$
$= (56 + 4x)$ or $4(14 + x)$ --- (1)
Semi perimeter of $\triangle$ ABC = $\frac{14+(6+x)+(8+x)}{2} = (14 + x)$
Also, area ($\triangle$ ABC) = $\sqrt{(14 + x)(14 + x – 14)[(14 + x) – (6 + x)][(14 + x) – (8 + x)]}$
$= \sqrt{48x(14 + x)}$ --- (2)
From (1) and (2), we get
$x=7$
$\therefore AB = 15$ cm and $AC = 13$ cm

(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).

(a) $20$ cm

(C) Rectangle

Join OP and OQ.
BR = BS = $24$ cm
$\therefore CR = 6$ cm
$\Rightarrow CQ = 6$ cm
Also, DQ = OP = $8$ cm
Hence, DC = $8 + 6 = 14$ cm

Join $OP, OQ, OR$ and $OS$
$\triangle POB \cong \triangle QOB$
$\Rightarrow \angle 1 = \angle 2$
Similarly $\angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$
Now, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$\Rightarrow 2(\angle 1 + \angle 8 + \angle 4 + \angle 5) = 360^\circ$
$\therefore \angle AOB + \angle COD = 180^\circ$

Correct figure
$\triangle OPA \cong \triangle OSA$
$\Rightarrow \angle 1 = \angle 2$
Similarly, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$
Now, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$
$\Rightarrow 2 (\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^{\circ}$
$\Rightarrow (\angle 1 + \angle 8) + (\angle 4 + \angle 5) = 180^{\circ}$
$\Rightarrow \angle AOD + \angle BOC = 180^{\circ}$

Correct figure. $\Delta OAP \cong \Delta OAS \implies \angle 1 = \angle 2$. Similarly, $\angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$. Also, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ \implies 2(\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^\circ \implies \angle AOB + \angle COD = 180^\circ$. Similarly, $\angle BOC + \angle AOD = 180^\circ$.

correct figure (1/2 Mark)
AM = AR
BM=BN
CN = CO
DO = DP
EQ=EP
FQ=FR (1 Mark)
LHS $= AB+CD+EF = (AM+BM) + (CO+OD) + (EQ + QF)$ (1/2 Mark)
$= AR+ BN + CN + DP + EP + FR$ (1/2 Mark)
$= (AR+FR) + (BN+CN)+ (DP+EP)$ (1/2 Mark)
$= BC + DE+ FA = \text{RHS}$

Let ABCD be a quadrilateral circumscribing a circle with centre O. Let the sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.
Join OA, OB, OC, OD, OP, OQ, OR, OS.
In $\triangle AOP$ and $\triangle AOS$:
OP = OS (Radii of the same circle)
OA = OA (Common side)
AP = AS (Tangents from an external point A)
So, $\triangle AOP \cong \triangle AOS$ (SSS congruence criterion)
$\Rightarrow \angle 1 = \angle 2$ (CPCTC)
Similarly, we can prove:
$\triangle BOQ \cong \triangle BOP \Rightarrow \angle 3 = \angle 4$
$\triangle COR \cong \triangle COQ \Rightarrow \angle 5 = \angle 6$
$\triangle DOR \cong \triangle DOS \Rightarrow \angle 7 = \angle 8$
The sum of all angles around the centre O is $360^\circ$.
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ$ (since $\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8$)
$\angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$
Now, consider angles subtended by opposite sides at the centre:
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8)$
Since $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$
$\angle AOB + \angle COD = (\angle 1 + \angle 3) + (\angle 5 + \angle 7)$ (This step is incorrect in the provided solution, it should be $\angle 1+\angle 4$ and $\angle 5+\angle 8$)
Let's re-evaluate: $\angle AOB = \angle 1 + \angle 4$, $\angle COD = \angle 5 + \angle 8$
$\angle BOC = \angle 3 + \angle 6$, $\angle DOA = \angle 2 + \angle 7$
We know $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$.
So, $2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ \Rightarrow \angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$.
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8) = (\angle 1 + \angle 3) + (\angle 5 + \angle 7) = 180^\circ$.
Similarly, $\angle BOC + \angle DOA = (\angle 3 + \angle 6) + (\angle 2 + \angle 7) = (\angle 3 + \angle 5) + (\angle 1 + \angle 7) = 180^\circ$.
Thus, opposite sides subtend supplementary angles at the centre.

ABCD is a parallelogram touching the circle at P, Q, R, S by sides AB, BC, CD, DA respectively.
We know that tangents drawn from the external point to a circle are equal.
$\therefore AP = AS$
quad --------(i)
$PB = BQ$
quad --------(ii)
$CR = CQ$
quad --------(iii)
$DR = DS$
quad ---------(iv)
Adding (i), (ii), (iii), (iv)
$(AP + PB) + (CR + DR) = (AS + DS) + (BQ + CQ)$
$AB + CD = AD + BC$
ABCD is a parallelogram
$\Rightarrow AB = CD, AD = BC$
$\Rightarrow 2AB = 2AD \Rightarrow AB = AD$
$\Rightarrow \text{ABCD is a rhombus.}$

$\angle OAP = 90^\circ$. In right $\Delta OAP, (26)^2 = OA^2 + (10)^2 \implies OA = \sqrt{576} = 24$. $\therefore \text{radius} = 24$ m.

(i) AR = $x$ m
(ii) Quad. ORBQ is a square.
(iii) (a) PC = $8 + x$
AC$^2 = (8 + 2x)^2 =49 + 225 =274$
$\Rightarrow 8 + 2x = \sqrt{274}$
$x = \frac{-8+\sqrt{274}}{2}$ or $4.28$ approx.
OR
(iii) (b) AC$^2 = (8 + 2x)^2 =49 + 225 =274$
$8 + 2x = \sqrt{274}$
$x = \frac{-8+\sqrt{274}}{2}$ or $4.28$ approx.
Hence, radius $r= 7 - x = 7 - \left(\frac{-4 + \sqrt{274}}{2}\right)$
$= \left(11 - \frac{\sqrt{274}}{2}\right)$ or $2.72$ approx.
Therefore, radius of the circle is $\left(11 - \frac{\sqrt{274}}{2}\right)$ m or $2.72$ m approx.

(i) Central angle $= \frac{360^\circ}{10} = 36^\circ$. (ii) Length of arc $ACB = \frac{1}{10} \times 2 \times \frac{22}{7} \times \frac{35}{2} = 11$ mm. (iii)(a) Area $= \frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 96.25$ mm$^2$. (iii)(b) Length of wire $= 2 \times \frac{22}{7} \times \frac{35}{2} + 5 \times 35 = 285$ mm.