Rd Sharma 2022 _mcqs for Class 10 Maths Chapter 8 - Circles

Answer:

Let us first put the given data in the form of a diagram.

Given data is as follows:

OQ = 12 cm

OP = 5 cm

We have to find the length of QP.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, OP is perpendicular to QP. We can now use Pythagoras theorem to find the length of QP.

Therefore the correct answer is choice (d).

Answer:

Let us first put the given data in the form of a diagram.

The given data is as follows:

QP = 24 cm

QO = 25 cm

We have to find the length of OP, which is the radius of the circle.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, OP is perpendicular to QP. We can now use Pythagoras theorem to find the length of QP.

Therefore the length of the radius of the circle is 7 cm.

Hence the correct answer to the question is choice (a).

Answer:

Let us first put the given data in the form of a diagram.

Given data is as follows:

OP = 3 cm

AP = 4 cm

We have to find the length of OA.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, OP is perpendicular to AP. We can now use Pythagoras theorem to find the length of OA.

Therefore, the distance of the point A from the center of the circle is 5 cm.

Hence the correct answer is choice (c).

Answer:

A circle can have two parallel tangents.

Hence, the correct answer is option (b).

Answer:

Given that, angle between the radii of a circle is 100°


From the figure, $\angle \mathrm{AOB}=100°$
Now, $\angle \mathrm{OAC}=90° \mathrm{and} \angle \mathrm{OBC}=90° \left(\mathrm{Radii} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{tangent} \mathrm{at} \mathrm{point} \mathrm{of} \mathrm{contact}\right)$

By angle sum property of a quadrilateral, the sum of interior angles of quadrilateral = 360°
⇒ 100° + 90° + 90° + $\angle \mathrm{ACB}$ =360°
$\Rightarrow \angle ACB=80°$

Hence, the correct answer is option (c).

Answer:

Let us first put the given data in the form of a diagram.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, OP is perpendicular to QP. Therefore,

The side opposite tois OQ.OQ will be the longest side of the triangle. So, in the isosceles right triangle ΔOPQ,

OP = PQ

And the angles opposite to these two sides will also be equal. Therefore,

We know that sum of all angles of a triangle will always be equal to 180°. Therefore,

Therefore, the correct answer to this question is choice (b).

Answer:

Let us first put the given data in the form of a diagram.

Let us draw radius OA, such that it touches the circle with center O at point A.

Let us draw radius O’B such that it touches the circle with center O’ at point B.

Let us also draw radii from each circle to the point C.

We know that the radius is always perpendicular to the tangent at the point of contact. Therefore,

That is,

…… (1)

Similarly,

Now, in ,

OA = OC(Radii of the same circle)

Therefore,

(Angles opposite to equal sides will be equal) …… (3)

Similarly, in

…… (4)

Consider the straight line OCO’. We have,

Looking at the figure, we can rewrite the above equation as,

(From (3) and (4))

(From (1) and (2))

…… (5)

Now let us take up. We have,

(Sum of all angles of a triangle will be 180°)

From equation (5), we get

Therefore the correct answer to this question is choice (d).

Answer:

Let us first put the given data in the form of a diagram.

Let us first find out AC using Pythagoras theorem.

Also, we know that tangents drawn from an external point will be equal in length. Therefore we have the following,

BL = BM …… (1)

CM = CN …… (2)

AL = AN …… (3)

We have found that,

AC = 10

That is,

AN + NC = 10

But from (2) and (3), we can say

AL + MC = 10 …… (4)

It is given that,

AB = 8

BC = 6

Therefore,

AB + BC =14

Looking at the figure, we can rewrite this as,

AL + LB + BM + MC = 14

(AL + MC) + (BM + BL) = 14

Using (1) and (3) we can write the above equation as,

10 + 2BL = 14

BL = 2Consider the quadrilateral, BLOM, we have,

BL = BM (From (1))

OL = OM(Radii of the same circle)

(Given data)

(Radii is always perpendicular to the tangent at the point of contact)

(Radii is always perpendicular to the tangent at the point of contact)

Since the sum of all angles of a quadrilateral will be equal to , we have,

Since all the angles of the quadrilateral are equal to and since adjacent sides are equal, the quadrilateral BLOM is a square. We know that all the sides of a square are of equal length.

We have found BL = 2

Therefore,

OM = 2

OM is nothing but the radius of the circle.

Therefore, radius of the circle is 2 cm.

Hence the correct answer to the question is option (b).

Answer:

Let us first put the given data in the form of a diagram.

Given data is as follows:

QOR is the diameter.

=

We have to find .

Since QOR is the diameter of the circle, it is a straight line. Therefore,

That is,

But = . Therefore,

Now consider . We have

(Sum of all angles of a triangle will be )

But,

(Since radius will be perpendicular to the tangent at the point of contact)

Therefore,

Therefore, the correct answer to this question is option (c).

Answer:

The figure of the Quadrilateral is drawn below.

We know that length of the tangents drawn from an external point will be equal. Therefore,

AP = PQ

DP = DS

CR = CS

BR = BQ

Let us add all the above four equations. We get

AP + DP + CR + BR = PQ + DS + CS + BQ

By looking at the figure, we can rewrite the above equation as,

AD + BC = AB + CD

Therefore, the correct answer is option (c).

Answer:

Let us first put the given data in the form of a diagram.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, OP is perpendicular to QP. We can now use Pythagoras theorem to find the length of QP.

Therefore, the correct answer is option (b).

Answer:

Let us first put the given data in the form of a diagram.

We know that tangents drawn from an external point will be equal.

Therefore,

AD = CD

Since CD is given as 4 cm,

AD = 4 cm

Similarly,

BD = CD

Therefore,

BD = 4 cm

AB = AD + BD

AB = 4 + 4

AB = 8 cm

Therefore, the answer to this question is option (c).

Answer:

Let PQ and RP be the radii of the circle with the centre O.
$\angle \mathrm{ROQ}=130°$
$\mathrm{RP}\perp \mathrm{OR} \mathrm{and} \mathrm{PQ}\perp \mathrm{OQ}$            (Radii are perpendicular to the tangent)
In quadrilateral ROQP, 
$$\begin{gathered} \angle \mathrm{ORP}+\angle \mathrm{RPQ}+\angle \mathrm{PQO}+\angle \mathrm{QOR}=360° \\ \Rightarrow 90°+\angle \mathrm{RPQ}+90°+130°=360° \\ \Rightarrow \angle \mathrm{RPQ}=50° \end{gathered}$$
Hence, the correct answer is option (b).
 

Answer:

The maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is 2.

Hence, the correct answer is option (b)

Answer:

Given that, two perpendicular tangents PA and PB are drawn from an external point to a circle of radius 4 cm.

From the figure,
CA is perpendicular to AP and CB is perpendicular to BP 
Also,
AC = BC (Radius of a circle)
AP = BP (Tangent to circle)

So, BPAC is a square.

Therefore, the length of the tangent each is 4 cm.

Hence, the correct answer is an option (b).

Answer:

We have, O'Q  is perpendicular to AT
Also, OT $\perp$ AT

In $△\mathrm{AQO}\text{'} \mathrm{and} △\mathrm{ATO}$ 
$$\begin{gathered} \angle \mathrm{AQO}\text{'}=\angle \mathrm{ATO} \\ \angle \mathrm{QAO}\text{'}=\angle \mathrm{TAO} \end{gathered}$$

By AA similarity, $△\mathrm{AQO}\text{'} \mathrm{and} △\mathrm{ATO}$ are similar to each other.
$\frac{\mathrm{AO}\text{'}}{\mathrm{AO}}=\frac{\mathrm{QO}\text{'}}{\mathrm{TO}} \left(\mathrm{Corresponding} \mathrm{parts} \mathrm{of} \mathrm{similar} \mathrm{triangles}\right)$

Now,
$$\begin{gathered} \mathrm{AO}=\mathrm{AO}\text{'}+\mathrm{O}\text{'}\mathrm{P}+\mathrm{OP} \\ =3\mathrm{AO}\text{'} (\mathrm{Radii} \mathrm{of} \mathrm{circle} \mathrm{are} \mathrm{equal}) \\ \Rightarrow \frac{\mathrm{AO}\text{'}}{\mathrm{AO}}=\frac{\mathrm{AO}\text{'}}{3\mathrm{AO}}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{AQ}}{\mathrm{AT}}=\frac{\mathrm{AO}\text{'}}{\mathrm{AO}}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{AQ}}{\mathrm{AT}}=\frac{1}{3} \end{gathered}$$

Hence, the correct answer is option (d).
 

Answer:

Given that, from a point A which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents AB and AC to the circle are drawn.



From the figure, we have AB and AC are the tangents to the circle which are perpendicular to the radius of the circle.
Thus,
$$\begin{gathered} \mathrm{AB}\perp \mathrm{OB} \mathrm{and} \mathrm{AC}\perp \mathrm{OC} \\ △\mathrm{ABO} \mathrm{and} △\mathrm{ACO} \mathrm{are} \mathrm{right}-\mathrm{angled} \mathrm{triangles} \end{gathered}$$

Using Pythagoras theorem in $△\mathrm{ABO}$
$$\begin{gathered} {\left(\mathrm{AB}\right)}^2+{\left(\mathrm{OB}\right)}^2={\left(\mathrm{AO}\right)}^2 \\ \Rightarrow {\left(\mathrm{AB}\right)}^2+5^2={13}^2 \\ \Rightarrow {\left(\mathrm{AB}\right)}^2+25=169 \\ \Rightarrow {\left(\mathrm{AB}\right)}^2=144 \\ \Rightarrow \mathrm{AB}=12 \mathrm{cm} \end{gathered}$$

We know, tangents to a circle from an external point are equal.
So, AB = AC = 12 cm

Therefore, Area of quadrilateral ABOC = Area of $△$AOB + Area of $△$AOC
                                                               $$\begin{gathered} =\left(\frac{1}{2}\times \mathrm{OB}\times \mathrm{AB}\right)+\left(\frac{1}{2}\times \mathrm{OC}\times \mathrm{AC}\right) \left[\because \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}bh\right] \\ =\left(\frac{1}{2}\times 5\times 12\right)+\left(\frac{1}{2}\times 5\times 12\right) \\ =60 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct answer is option (a).

 

Answer:

In quadrilateral APBO,
$\angle APB=50°$
$$\begin{gathered} \angle OAP+\angle APB+\angle PBO+\angle AOB=360° \\ \Rightarrow 90°+90°+50°+\angle AOB=360° \\ \Rightarrow \angle AOB=130° \end{gathered}$$
In ∆AOB,
Let $\angle OAB=\angle OBA=x$            (OA = OB as they are radii)
$$\begin{gathered} \angle AOB+\angle OBA+\angle OAB=180° \\ \Rightarrow 130°+2x=180 \\ \Rightarrow x=25° \end{gathered}$$
Hence, the correct answer is option (a).

Answer:

Since AB is the diameter of the circle, the angle subtended by it on the circumference is 90°.
Therefore, $\angle \mathrm{ACB}=90°$.

By angle sum property of triangle,
$\angle \mathrm{ABC}=60° \left(\because \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180° \right)$

Since tangent is perpendicular to the radius at the point of contact, we have
$\angle \mathrm{OCD}=90°$
$$\begin{gathered} \Rightarrow \angle \mathrm{OCB}+\angle \mathrm{BCD}=90° \\ \Rightarrow \angle \mathrm{OBC}+\angle \mathrm{BCD}=90° \left[\because \mathrm{angles} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{are} \mathrm{equal}\right] \\ \Rightarrow \angle \mathrm{BCD}=90°-60° \\ \Rightarrow \angle \mathrm{BCD}=30° \end{gathered}$$

In $△$BCD, we have by exterior angle property, 
$$\begin{gathered} \angle \mathrm{OBC}=\angle \mathrm{BCD}+\angle \mathrm{BDC} \\ \Rightarrow \angle \mathrm{BDC}=60°-30° \\ \Rightarrow \angle \mathrm{BDC}=30° \end{gathered}$$

So, in $△$BCD, we have $\angle \mathrm{BCD}=\angle \mathrm{BDC}=30°$
So, BC = BD. Thus, $△$BCD is isosceles.

Hence, the correct answer is option (c).

 

Answer:

By looking at the given figure

We can write the following equations:

AD + DB = 12

We know that tangents form an external point will be of equal length. Therefore,

DB = BE …… (1)

Hence we have,

AD + BE = 12

From the figure, we have

BE + EC = 8 …… (2)

Let us subtract equation (2) from equation (3). We get,

AD − EC = 4 …… (3)

Since tangents from an external point will be equal, we have,

AD = AF

EC = CF

Therefore, equation (3) becomes,

AF − CF = 4 …… (4)

From the figure we have,

AF + CF = 10 …… (5)

Adding equation (4) and (5), we get,

2AF = 14

AF = 7

We know that,

AF = AD

Therefore,
AD = 7

The correct answer is option (d).

Answer:

The given figure is below

It is given that, 

AP = PB

We know that

PB = BR(tangents from an external point are equal)

From the above two equations, we can say that,

AP = BR …… (1)

Also,

AP = AQ(tangents from an external point are equal) …… (2)

From equations (1) and (2), we have,

AQ = BR …… (3)

Also,

QC = CR(tangents from an external point are equal) …… (4)

Adding equations (3) and (4), we get,

AQ + QC = BR + CR

By looking at the figure we can write the above equation as,

AC = BC

The correct answer is option (b).

Answer:

Since the radius is always perpendicular to the tangent at the point of contact, 

Therefore,

Now, consider. Here also, OB is perpendicular to PB since the radius will be perpendicular to the tangent at the point of contact. Therefore,

The correct answer is option (b).

Answer:

We know that the radius is always perpendicular to the tangent at the point of contact.  

Therefore, we have,

It is given that,

That is,

Now, consider. We have,

OP = OQ (Radii of the same circle)

Since angles opposite to equal side will be equal in a triangle, we have,

We know that sum of all angles of a triangle will be equal to .

Therefore,

The correct answer is option (c).

Answer:

We know that tangents drawn to a circle from the same external point will be equal in length.

Therefore,

PD = PA …… (1)

QB = QA …… (2)

Adding equations (1) and (2), we get,

PD + QB = PA + QA

By looking at the figure we can say,

PD + QB = PQ

Therefore option (a) is the correct answer.

Answer:

We know that tangents drawn from the same external point will be equal in length. 

Therefore, we have,

QP = PT

PT = PR

From the above two equations, we get,

QP = PR

It is given that,

QP = 4.5 cm

Therefore,

PR = 4.5 cm

From the figure, we have,

QR = QP + PR

QR = 4.5 + 4.5

QR = 9

The correct answer is option (a).

Answer:

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact.

Therefore,

That is,

It is given that,

Therefore, we have,

Now, consider . We have,

OP = OQ(Radii of the same circle)

Since angles opposite to equal sides will be equal, we have,

We have found that,

Therefore,

We know that sum of all angles of a triangle will be equal to . Therefore,

The correct answer is option (a).

Answer:

Since tangents drawn from an external point to a circle are equal.
PA = PB                 .....(1)
QA = QD                 .....(2)
SC = SB                 .....(3)
RC = RD                 .....(4)



Adding the LHS and RHS of (1), (2), (3) and (4), we get
$$\begin{gathered} \left(\mathrm{PA}+\mathrm{QA}\right)+\left(\mathrm{SC}+\mathrm{RC}\right)=\left(\mathrm{PB}+\mathrm{SB}\right)+\left(\mathrm{DQ}+\mathrm{RD}\right) \\ \Rightarrow \mathrm{PQ}+\mathrm{RS}=\mathrm{PS}+\mathrm{QR} \\ \Rightarrow 6.5+\mathrm{RS}=4.2+7.3 \\ \Rightarrow \mathrm{RS}=5 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

We know that the radius will always be perpendicular to the tangent at the point of contact.

Therefore,

Hence we have,

Also,

OO’ = sum of the radii of the two circles

OO’ = 3 + 5

OO’ = 8

Since radius is perpendicular to the tangent, . Therefore,

PR = PO + OO’ + O’R

PR = 5 + 8 + 13

PR =26

Therefore, option (b) is correct.

Answer:

From the given figure we have,

AO = r + r + r

AO = 3r

AO’ = r

Therefore,

Also as therefore

Therefore, option (c) is correct.

Answer:

Consider .

We have,

Therefore,

Considering AB as the chord to the bigger circle, as OQ is perpendicular to AB, OQ bisects AB.

∴ AQ = QB = 4 cm.

Now, as BQ and BP are pair of tangents to the inner circle drawn from the external point B, QB = PB = 4 cm.

Now, join OP.

Then, _.

⇒ OP bisects BC

⇒ BP = PC = 4 cm

Thus, BC = 4 cm + 4 cm = 8 cm

Therefore, option (c) is correct.

Answer:

Let us first draw the radii OS, OP and OQ in the given figure, for our convenience.

Consider . We have,

PO = OS (Radii of the same circle)

Therefore, is an isosceles triangle.

We know that in an isosceles triangle, if a line is drawn perpendicular to the base of the triangle from the common vertex of the equal sides, then that line will bisect the base (unequal side).

Therefore, we have

PQ = QS

It is given that,

PQ = 7.5

Therefore,

QS = 7.5

From the figure, we have,

PS = PQ + QS

PS = 7.5 + 7.5

PS = 15

Therefore, option (c) is the correct answer.

Answer:

We know that tangents drawn from the same external point will be equal in length.

Therefore,

AB = AC

It is given that,

AB = 8 cm

Hence,

AC = 8 cm …… (1)

Similarly,

PE = CE

It is given that,

PE = 3

Therefore,

CE = 3 …… (2)

Subtracting equations (1) and (2), we get,

AC − CE = 8 − 3

From the figure we can see that,

AC − CE = AE

Therefore,

AE = 8 − 3

AE = 5 cm

Option (c) is the correct answer.

Answer:

Consider and . We have,

PO is the common side for both the triangles.

OQ = OR(Radii of the same circle)

PQ = PR(Tangents from an external point will be equal)

Therefore, from SSS postulate of congruent triangles, we have,

Therefore,

That is,

It is given that,

Therefore,

Now consider . We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,

We know that sum of all angles of a triangle will always be equal to . Therefore,

Therefore, the correct answer is option (b).

Answer:

Consider .

We have,

TP = TQ(Tangents from an external point will be equal)

We know that angles opposite to equal sides will be equal. Therefore,

It is given that,

Therefore,

We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore,

Hence,

That is,

We have found that,

Therefore,

Therefore, the correct answer is option (b).

Answer:

It is given that the sides AB , BC and CA of ∆ABC touch a circle at P, Q and R, respectively.

Also, PA = 4 cm, PB = 3 cm and AC = 11 cm

We know that, the lengths of tangents drawn from an external point to a circle are equal.

∴ AR = AP = 4 cm

BQ = BP = 3 cm

Now, CR = AC − AR = 11 cm − 4 cm = 7 cm

∴ CQ = CR = 7 cm         (Lengths of tangents drawn from an external point to a circle are equal)

Now,

BC = BQ + CQ = 3 cm + 7 cm = 10 cm

Thus, the length of BC is 10 cm.

Hence, the correct answer is option B.            

Answer:

In the given figure, DH and DK are tangents drawn to the circle from an external point D.

∴ DH = DK                      (Lengths of tangents drawn from an external point to a circle are equal)

FH and FM are tangents drawn to the circle from an external point F.

∴ FH = FM                      (Lengths of tangents drawn from an external point to a circle are equal)

EK and EM are tangents drawn to the circle from an external point E.

∴ EM = EK = 9 cm          (Lengths of tangents drawn from an external point to a circle are equal)

Now,

Perimeter of ΔEDF = ED + DF + EF

                           = ED + (DH + FH) + EF

                           = ED + (DK + FM) + EF                (DH = DK and FH = FM)

                           = (ED + DK) + (FM + EF)

                           = EK + EM

                           = 9 cm + 9 cm

                           = 18 cm

Thus, the perimeter of ΔEDF is 18 cm.

Hence, the corrrect answer is option A.

Answer:

It is given that DE and DF are tangents from an external point D to a circle with centre A. DE = 5 cm and DE ⊥ DF.



Join AE and AF.

Now, DE is a tangent at E and AE is the radius through the point of contact E.

$\therefore \angle \mathrm{AED}=90°$     (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Also, DF is a tangent at F and AF is the radius through the point of contact F.

$\therefore \angle \mathrm{AFD}=90°$     (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$\angle \mathrm{EDF}=90°$         (DE ⊥ DF)

Also, DF = DE        (Lengths of tangents drawn from an external point to a circle are equal)

So, AEDF is a square.

∴ AE = AF = DE = 5 cm         (Sides of square are equal)

Thus, the radius of the circle is 5 cm.

Hence, the correct answer is option B.

Answer:

It is given that the sides BC, AB, AD and CD touches a circle with centre O at points P, Q, R and S, respectively.

AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm.

Now, 

DR = DS = 5 cm           (Lengths of tangents drawn from an external point to a circle are equal)

AR = AD − DR = 23 cm − 5 cm = 18 cm

∴ AQ = AR = 18 cm           (Lengths of tangents drawn from an external point to a circle are equal)

BQ = AB − AQ = 29 cm − 18 cm = 11 cm

∴ BP = BQ = 11 cm           (Lengths of tangents drawn from an external point to a circle are equal)

Now, AB is a tangent at Q and OQ is the radius through the point of contact Q.

$\therefore \angle \mathrm{OQB}=90°$     (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

BC is a tangent at P and OP is the radius through the point of contact P.

$\therefore \angle \mathrm{OPB}=90°$     (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$\angle \mathrm{B}=90°$      (Given)

Also, BP = BQ

So, OPBQ is a square.

∴ OP = OQ = BQ = 11 cm         (Sides of square are equal)

Thus, the radius of the circle is 11 cm.

Hence, the correct answer is option A.

Answer:

Let a circle with centre O and radius r cm be inscribed in the right ∆ABC.

In right ∆ABC,

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^{2 } \left(\mathrm{Pythagoras} \mathrm{theorem}\right) \\ \Rightarrow \mathrm{AC}=\sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2} \\ \Rightarrow \mathrm{AC}=\sqrt{5^2+{12}^2} \\ \Rightarrow \mathrm{AC}=\sqrt{169}=13 \mathrm{cm} \end{gathered}$$

Now, the sides AB, BC and CA are tangents to the circle at R, P and Q, respectively.

∴ OR ⊥ AB, OP ⊥ BC and OQ ⊥ AC

Now,

ar(∆OAB) + ar(∆OBC) + ar(∆OAC) = ar(∆ABC)

$$\begin{gathered} \Rightarrow \frac{1}{2}\times \mathrm{AB}\times \mathrm{OR}+\frac{1}{2}\times \mathrm{BC}\times \mathrm{OP}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OQ}=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB} \\ \Rightarrow \frac{1}{2}\times 5\times r+\frac{1}{2}\times 12\times r+\frac{1}{2}\times 13\times r=\frac{1}{2}\times 12\times 5 \\ \Rightarrow 5r+12r+13r=60 \\ \Rightarrow 30r=60 \\ \Rightarrow r=2 \end{gathered}$$

Thus, the radius of the circle is 2 cm.

Hence, the correct answer is option C.

Answer:

It is given that two circles touch each other externally at P. AB is a common tangent to the circles touching them at A and B.



Draw a tangent to the circles at P, intersecting AB at T.

Now, TA and TP are tangents drawn to the same circle from an external point T.

∴ TA = TP            (Lengths of tangents drawn from an external point to a circle are equal)

TB and TP are tangents drawn to the same circle from an external point T.

∴ TB = TP            (Lengths of tangents drawn from an external point to a circle are equal)

In ∆ATP,

TA = TP

$\therefore \angle \mathrm{APT}=\angle \mathrm{PAT}$      .....(1)           (In a triangle, equal sides have equal angles opposite to them)

In ∆BTP,

TB = TP

$\therefore \angle \mathrm{BPT}=\angle \mathrm{PBT}$      .....(2)            (In a triangle, equal sides have equal angles opposite to them)

Now, in ∆APB,

$\angle \mathrm{APB}+\angle \mathrm{PAB}+\angle \mathrm{PBA}=180°$       (Angle sum property)
$$\begin{gathered} \Rightarrow \angle \mathrm{APB}+\angle \mathrm{APT}+\angle \mathrm{BPT}=180° \left[\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right] \\ \Rightarrow \angle \mathrm{APB}+\angle \mathrm{APB}=180° \\ \Rightarrow 2\angle \mathrm{APB}=180° \\ \Rightarrow \angle \mathrm{APB}=90° \end{gathered}$$

Thus, the value of $\angle$APB is 90º.

Hence, the correct answer is option D.

Answer:

It is given that PQ and PR are two tangents to a circle with centre O.

Now, PQ is a tangent at Q and OQ is the radius through the point of contact Q.

∴ $\angle$OQP = 90º               (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

PR is a tangent at R and OR is the radius through the point of contact R.

∴ $\angle$ORP = 90º               (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Also, $\angle$QPR = 46º    (Given)

In quadrilateral OQPR,

$\angle$OQP + $\angle$QPR + $\angle$ORP + $\angle$QOR = 360º                  (Angle sum property)

⇒ 90º + 46º + 90º + $\angle$QOR = 360º 

⇒ 226º + $\angle$QOR = 360º 

⇒ $\angle$QOR = 360º − 226º = 134º

∴ $\angle$QOR = 134º

Hence, the correct answer is option B.

Answer:

It is given that QR is a common tangent to the given circles touching externally at the point T. Also, the tangent at T meets QR at P such that PT = 3.8 cm.
 
Now, PQ and PT are tangents drawn to the same circle from an external point P.

∴ PQ = PT = 3.8 cm            (Lengths of tangents drawn from an external point to a circle are equal)

PR and PT are tangents drawn to the same circle from an external point T.

∴ PR = PT = 3.8 cm            (Lengths of tangents drawn from an external point to a circle are equal)

Now,

QR = PQ + PR = 3.8 cm + 3.8 cm = 7.6 cm

Thus, the length of QR is 7.6 cm.

Hence, the correct answer is option B.

Answer:

It is given that a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S, respectively.

Also, BC = 7 cm, CR = 3 cm and AS = 5 cm

Now, AP and AS are tangents drawn to the same circle from an external point A.

∴ AP = AS = 5 cm          .....(1)                (Lengths of tangents drawn from an external point to a circle are equal)

CQ and CR are tangents drawn to the same circle from an external point C.

∴ CQ = CR = 3 cm           (Lengths of tangents drawn from an external point to a circle are equal)

So, BQ = BC − CQ = 7 cm − 3 cm = 4 cm

BP and BQ are tangents drawn to the same circle from an external point B.

BP = BQ = 4 cm          .....(2)                    (Lengths of tangents drawn from an external point to a circle are equal)

Now,

AB = AP + PB = 5 cm + 4 cm = 9 cm            [From (1) and (2)]

∴ x = 9              (AB = x cm)

Hence, the correct answer is option B.

Answer:

In the given problem, the Right Hand Side of all the options is same, that is,

AB + BC + CA

So, we shall find out AB + BC + CA and check which of the options has the Left Hand Side value which we will arrive at.

By looking at the figure, we can write,

AB + BC + CA = AB + BF + FC + CD

We know that tangents drawn from an external point will be equal in length. Therefore,

BF = BE

FC= CD

Now we have,

AB + BC + CA = AB + BE + CD + CA

AB + BC + CD = (AB + BE) + (CD + CA)

By looking at the figure, we write the above equation as,

AB + BC + CD = AE + AD

Since tangents drawn from an external point will be equal,

AE = AD

Therefore,

AB + BC + CD = AD + AD

AB + BC + CD = 2AD

Therefore option (b) is the correct answer.

Answer:

It is given that,

SQ = 6 cm

Since SQ passes through the centre of the circle O, it is the diameter. Therefore, the radius,

OQ = 3 cm

Also, given is

QR = 4

We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore, OQ is perpendicular to OR. We can find the length of OR by using Pythagoras theorem. We have,

Therefore option (d) is the correct answer.

Answer:

We know that tangents from an external point will be equal in length.  

Therefore,

AQ = AR

AQ is given as 4 cm. Therefore,

AR = 4

Similarly,

PC = CQ

PC = 5

Therefore,

CQ = 5

Similarly,

BP = BR

BR = 6

Therefore,

BP = 6

Now we can find out the perimeter of the triangle.

Perimeter = AB + BC + CA

From the figure we have,

Perimeter = AR + RB + BP + PC + CQ + QA

Perimeter = 4 + 6 + 6 + 5 + 5 + 4

Perimeter = 30 cm

Therefore, option (a) is the correct answer.

Answer:

In the given figure,

if we join O and A, the line OA will be perpendicular to AP. This is because the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, is a right triangle.

We know that,

…… (1)

We know that,

…… (2)

From equations (1) and (2), we have,

Therefore, option (c) is the correct answer.

Answer:

Let us first put the given data in the form of a diagram.

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore,

Since tangents drawn from an external point will be equal in length,

AP = AQ

Since, AP = 12

AQ = 12

AP + AQ = 12 + 12

AP + AQ = 24

Therefore option (c) is the correct answer to this question.

Answer:

Consider the figure.
fIG

In right angled $∆\mathrm{OMB}$,

By pythagoras theorem.

$$\begin{gathered} {\mathrm{OB}}^2={\mathrm{MB}}^2+{\mathrm{OM}}^2 \\ \Rightarrow 5^2={\mathrm{MB}}^2+3^2 \\ \Rightarrow 25-9={\mathrm{MB}}^2 \\ \Rightarrow 16={\mathrm{MB}}^2 \\ \Rightarrow 4=\mathrm{MB} \end{gathered}$$
Therefore, AM = 2MB = 2 × 4 = 8 cm.
Hence, the correct answer is option (d).

Answer:

Let us first put the given data in the form of a diagram.

We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore,

Consider . We know that sum of all angles of a triangle will be . Therefore,

Since , we have,

Choice (c) is the right answer.

Answer:

As in the given figure PQRS is a quadrilateral circumscribing the circle and we know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Answer:

Since tangents drawn from an external point to a circle are equal.
$$\begin{gathered} \Rightarrow \mathrm{AP}=\mathrm{AS}=2 \mathrm{cm} \\ \Rightarrow \mathrm{BP}=\mathrm{QB}=4 \mathrm{cm} \\ \Rightarrow \mathrm{DR}=\mathrm{DS}=5 \mathrm{cm} \\ \Rightarrow \mathrm{CR}=\mathrm{CQ}=6 \mathrm{cm} \left(\because \mathrm{CQ}-\mathrm{BQ}=6 \mathrm{cm}\right) \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Perimeter} \mathrm{of} \mathrm{quadrilateral} \mathrm{ABCD}=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA} \\ =\left(\mathrm{AP}+\mathrm{BP}\right)+\left(\mathrm{BQ}+\mathrm{QC}\right)+\left(\mathrm{CR}+\mathrm{DR}\right)+\left(\mathrm{DS}+\mathrm{AS}\right) \\ =2\mathrm{AP}+2\mathrm{BP}+2\mathrm{DS}+2\mathrm{CQ} \\ =4+8+10+12 \\ =34 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Since tangents drawn from an external point to a circle are equal.
$\Rightarrow \mathrm{QA}=\mathrm{QB}=4 \mathrm{cm}$
Now, 
$\begin{array}{rcl}\mathrm{RB}& =& \mathrm{RQ}-\mathrm{BQ} \\ & =& 7- 4\\ & =& 3 \mathrm{cm}\\ & & \end{array}$

$$\begin{gathered} \mathrm{Since} \mathrm{tangents} \mathrm{drawn} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{to} \mathrm{a} \mathrm{circle} \mathrm{are} \mathrm{equal}. \\ \therefore \mathrm{RB}=\mathrm{RC}=3 \mathrm{cm} \mathrm{and} \mathrm{SD}=\mathrm{SC}=3 \mathrm{cm} \end{gathered}$$

Now,
$$\begin{gathered} \mathrm{RS}=x=\mathrm{SC}+\mathrm{RC} \\ =3+3 \\ =6 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

Joining the center of circle O with point P. We have

Since tangents drawn from an external point are always equal.
$$\begin{gathered} \therefore \mathrm{AQ}=\mathrm{AP}=14 \mathrm{cm} \\ \mathrm{Also}, \angle \mathrm{QAP}=\angle \mathrm{OQA}=\angle \mathrm{APO}=90° \\ \Rightarrow \angle \mathrm{APO}=90° \\ \mathrm{Also}, \mathrm{pair} \mathrm{of} \mathrm{adjacent} \mathrm{sides} \mathrm{are} \mathrm{equal}. \\ \Rightarrow \mathrm{OQAP} \mathrm{is} \mathrm{a} \mathrm{square}. \end{gathered}$$

Similarly,
$$\begin{gathered} \Rightarrow \mathrm{CR}=\mathrm{CS}=23 \mathrm{cm} \\ \Rightarrow \mathrm{BS}=\mathrm{BP}=39-23=16 \mathrm{cm} \end{gathered}$$

Thus, AB = 14 + 16 = 30 cm

Hence, the correct answer is option (d).
 

Answer:

Given that the two circles touch each other externally at P. AB is the common tangent to the two circles.

                   
In $△\mathrm{PAC}$, 
CA = CP    (∵ Length of the tangent from the external point of a circle are equal.)     
$\angle \mathrm{CAP}=\angle \mathrm{PAC}=\theta \left(\because \mathrm{In} \mathrm{a} \mathrm{triangle} \mathrm{angles} \mathrm{corresponding} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{also} \mathrm{equal}.\right)$

Similarly, in a $△\mathrm{PBC}$,
CB = CP   (∵ Length of the tangent from the external point of a circle are equal.)    
 $\angle \mathrm{CPB}=\angle \mathrm{CBP}=\phi \left(\because \mathrm{In} \mathrm{a} \mathrm{triangle} \mathrm{angles} \mathrm{corresponding} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{also} \mathrm{equal}.\right)$

In $△\mathrm{PAB}$, by using Angle Sum Property of triangle

$\angle \mathrm{PAB}+\angle \mathrm{PBA}+\angle \mathrm{APB}=180°$
$$\begin{gathered} \Rightarrow \theta +\phi +\left(\theta +\phi \right)=180° \\ \Rightarrow 2\left(\theta +\phi \right)=180° \\ \Rightarrow \left(\theta +\phi \right)=90° \\ \Rightarrow \angle \mathrm{APB}=90° \end{gathered}$$

Hence, the correct answer is option (d).

      

Answer:

Since the tangent to a circle is perpendicular to the radius. 
$\therefore \mathrm{OC}\perp \mathrm{BC}$
By Pythagoras theorem, we have
$$\begin{gathered} {\left(\mathrm{OB}\right)}^2={\left(\mathrm{OB}\right)}^2+{\left(\mathrm{BC}\right)}^2 \\ {15}^2=9^2+{\left(\mathrm{BC}\right)}^2 \\ 225 =81+{\left(\mathrm{BC}\right)}^2 \\ {\left(\mathrm{BC}\right)}^2=144 \\ \Rightarrow \mathrm{BC}=12 \mathrm{cm} \end{gathered}$$

Now, tangents to a circle from an external point are equal.
Therefore BC + BD = 12 + 12
                                 = 24 cm

Hence, the correct answer is option (c).

 

Answer:

 (i) The tangents to the Moon's surface from point A are AB and AC.

Hence, the correct answer is option (a).

(ii) Given that, ∠PAQ = 40°
As the tangent to a circle is perpendicular to the radius through the point of contact.
Here, OP⊥ PA and OQ⊥ QA
$\therefore \angle \mathrm{OPA}=\angle \mathrm{OQA}=90°$

Now, in quadrilateral POAQ,

$$\begin{gathered} \angle \mathrm{POQ}+\angle \mathrm{OPA}+\angle \mathrm{PAQ}+\angle \mathrm{OQA}=360° \\ \Rightarrow \angle \mathrm{POQ}+\angle \mathrm{PAQ}=180° \left(\because \angle \mathrm{OPA}=\angle \mathrm{OQA}=90° \right) \\ \Rightarrow \angle \mathrm{POQ}+40°=180° \\ \Rightarrow \angle \mathrm{POQ}=140° \end{gathered}$$

Hence, the correct answer is option (d).

(iii) Given that, ∠POQ = 110°
In $△\mathrm{POA} \mathrm{and} △\mathrm{QOA}$
$$\begin{gathered} \mathrm{PO}=\mathrm{OQ} \left(\mathrm{Radii} \mathrm{of} \mathrm{same} \mathrm{circle}\right) \\ \angle \mathrm{OPA}=\angle \mathrm{OQA} \left(\mathrm{Each} 90°\right) \\ \mathrm{OA}=\mathrm{AO} \left(\mathrm{Common}\right) \\ \therefore △\mathrm{POA} \cong △\mathrm{QOA} \left(\mathrm{By} \mathrm{RHS} \mathrm{congruency}\right) \\ \Rightarrow \angle \mathrm{POA}=\angle \mathrm{QOA} \left(\mathrm{By} \mathrm{CPCT}\right) \\ \Rightarrow \angle \mathrm{POA}=\angle \mathrm{QOA} =\frac{\angle \mathrm{PO}Q}{2}=\frac{110°}{2}=55° \end{gathered}$$
In $△\mathrm{QOA}$
$$\begin{gathered} \angle \mathrm{AOQ}+\angle \mathrm{AQO}+\angle \mathrm{OAQ}=180° \\ \Rightarrow 55°+90°+\angle \mathrm{OAQ}=180° \\ \Rightarrow \angle \mathrm{OAQ}=35° \end{gathered}$$

Hence, the correct answer is an option (b).

(iv) In $△\mathrm{POQ}$, 
OP = OQ (Radii of the same circle)
∴ $\angle \mathrm{OQP}=\angle \mathrm{OPQ}$    (∵ In a triangle angles corresponding to equal sides are also equal)

In $△\mathrm{POQ}$, by using Angle Sum Property of triangle, we get
$$\begin{gathered} \angle \mathrm{POQ}+\angle \mathrm{OQP}+\angle \mathrm{OPQ}=180° \\ \Rightarrow 130°+2\angle \mathrm{OPQ}=180° \\ \Rightarrow 2\angle \mathrm{OPQ}=50° \\ \Rightarrow \angle \mathrm{OPQ}=25° \end{gathered}$$

Hence, the correct answer is option (d).

(v) Given that,
OP = 9 units and OA = 41 units
Since in $△\mathrm{POA}$, $\angle P$ = 90$°$
$$\begin{gathered} \Rightarrow {\left(\mathrm{OA}\right)}^2={\left(\mathrm{OP}\right)}^2+{\left(\mathrm{PA}\right)}^2 \\ \Rightarrow {41}^2=9^2+{\left(\mathrm{PA}\right)}^2 \\ \Rightarrow 1681=81+{\left(\mathrm{PA}\right)}^2 \\ \Rightarrow {\left(\mathrm{PA}\right)}^2=1600 \\ \Rightarrow \left(\mathrm{PA}\right)=40 \mathrm{cm} \end{gathered}$$
Since the length of tangent drawn from an external point are equal.
Thus, AP = AQ = 40 cm

Hence, the correct answer is option (c).

 





 

Answer:

(i) Given that,  $\mathrm{Ref}\angle \mathrm{IOA}=240°$
$\Rightarrow \angle \mathrm{IOA}=120° \left(\because \mathrm{Complete} \mathrm{angle}=360°\right)$

Since the tangent to a circle is perpendicular to the radius through the point of contact.
Here, PI ⊥ IO and PA ⊥ AO
$\Rightarrow \angle \mathrm{OIP}=90° \mathrm{and} \angle \mathrm{OAP}=90°$

In quadrilateral CIOA, 
$$\begin{gathered} \Rightarrow \angle \mathrm{IOA}+\angle \mathrm{OAP}+\angle \mathrm{API}+\angle \mathrm{PIO}=360° \left(\because \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 360° \right) \\ \Rightarrow 120°+90°+\angle \mathrm{IPA}+90°=180° \\ \Rightarrow \angle \mathrm{IPA}=60° \end{gathered}$$

Hence, the correct answer is an option (c).

(ii) Given that, IP = 15 cm

Here, join OP.
$$\begin{gathered} \mathrm{In} △\mathrm{IOP} \mathrm{and}△\mathrm{AOP} \\ \mathrm{OI}=\mathrm{OA} \left(\mathrm{Radii} \mathrm{of} \mathrm{the} \mathrm{same} \mathrm{circle}\right) \\ \mathrm{IP}=\mathrm{AP} \left(\mathrm{The} \mathrm{lengths} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{tangents} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{to} \mathrm{a} \mathrm{circle} \mathrm{are} \mathrm{equal}.\right) \\ \mathrm{OP}=\mathrm{OP} \left(\mathrm{Common}\right) \\ \Rightarrow △\mathrm{IOP} \cong △\mathrm{AOP} \left(\mathrm{By} \mathrm{SSS} \mathrm{congruency}\right) \\ \Rightarrow \angle \mathrm{IOP}=\angle \mathrm{AOP}=60° \left(\mathrm{By} \mathrm{CPCT}\right) \\ \mathrm{And} \angle \mathrm{IPO}=\angle \mathrm{APO} \left(\mathrm{By} \mathrm{CPCT}\right) \end{gathered}$$

$\Rightarrow \angle \mathrm{IOB}=\angle \mathrm{AOB}=60°$
Now,
$$\begin{gathered} \mathrm{In} △\mathrm{OIP} \\ \tan 60°=\frac{\mathrm{PI}}{\mathrm{OI}} \\ \Rightarrow \sqrt{3}=\frac{15}{\mathrm{OI}} \\ \Rightarrow \mathrm{OI}=5\sqrt{3} \mathrm{cm} \end{gathered}$$
Therefore, the radius of the circle is $5\sqrt{3} \mathrm{cm}$.

$$\begin{gathered} \mathrm{In} △\mathrm{IBO} \mathrm{and}△\mathrm{ABO} \\ \mathrm{OI}=\mathrm{OA} \left(\mathrm{Radii} \mathrm{of} \mathrm{the} \mathrm{same} \mathrm{circle}\right) \\ \angle \mathrm{IOB}=\angle \mathrm{AOB} \left(\mathrm{Each} 60°\right) \\ \mathrm{OB}=\mathrm{OB} \left(\mathrm{Common}\right) \\ \Rightarrow △\mathrm{IBO} \cong △\mathrm{ABO} \left(\mathrm{By} \mathrm{SAS} \mathrm{congruency}\right) \\ \Rightarrow \angle \mathrm{IBO}=\angle \mathrm{ABO}=90° \left(\mathrm{By} \mathrm{CPCT}\right) \\ \mathrm{And} \mathrm{IB}=\mathrm{IA} \left(\mathrm{By} \mathrm{CPCT}\right) \end{gathered}$$

Also,
$$\begin{gathered} \mathrm{In} △\mathrm{OIB} \\ \sin 60°=\frac{\mathrm{IB}}{\mathrm{OI}} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{IB}}{5\sqrt{3}} \\ \Rightarrow \mathrm{IB}=7.5 \mathrm{cm} \end{gathered}$$
Now, OB is the line bisector of AI
$$\begin{gathered} \therefore \mathrm{AI}=2\times \mathrm{IB} \\ =2\times 7.5 \\ =15 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is an option (b).

(iii) Given that, IP = 21 cm and AP = x² + 5
Now, tangents drawn from a point are equal in length.
$$\begin{gathered} \therefore 21=x^2+5 \\ \Rightarrow x^2=16 \\ \Rightarrow x=4 \mathrm{cm} \end{gathered}$$
Thus, the value of x is 4 cm.

Hence, the correct answer is option (a).

(iv) We have,
$\angle \mathrm{OIP}=90° \left(\because \mathrm{Tangent} \mathrm{to} \mathrm{a} \mathrm{circle} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{the} \mathrm{radius} \mathrm{through} \mathrm{the} \mathrm{point} \mathrm{of} \mathrm{contact}\right)$
Also, $\angle \mathrm{IPA}=60° \left(\mathrm{From} \left(\mathrm{i}\right)\right)$
$\Rightarrow \angle \mathrm{IPO}=\angle \mathrm{APO}=30° \left(\because \mathrm{OP} \mathrm{is} \mathrm{angle} \mathrm{bisector} \mathrm{of} \angle \mathrm{IPA} \mathrm{and} \angle \mathrm{IOA}\right)$
$\begin{array}{rcl}\therefore \angle \mathrm{OIP}+\angle \mathrm{APO}& =& 90°+30°\\ & =& 120°\end{array}$
 Hence, the correct answer is option (c).

(v) The angles subtended at the center by the arcs  $\stackrel{⏜}{\mathrm{IZA}} \mathrm{and} \stackrel{⏜}{\mathrm{IXA}}$ are 240º and 120º.
$\begin{array}{rcl}\therefore \frac{\stackrel{⏜}{\mathrm{IZA}}}{\stackrel{⏜}{\mathrm{IXA}}}& =& \frac{{\displaystyle \frac{240°}{360°}}\times 2\mathrm{\pi }r}{\frac{120°}{360°}\times 2\mathrm{\pi }r}\\ & =& \frac{240°}{120°}\\ & =& \frac{2}{1}\end{array}$
Hence, the correct answer is option (c).

Answer:

Statement-2: A tangent to a circle is perpendicular to the radius through the point of contact.

Thus, Statement-2 is true.

Statement-1 (Assertion): The length of the tangent drawn from a point at a distance of 13 cm from the centre of a circle of radius 5 cm is 10 cm.

Given that,
OA = 13 cm and OP = radius of circle = 5 cm

Here, OP ⊥ PA
$\Rightarrow \angle \mathrm{OPA}=90°$       (∵The tangent to a circle is perpendicular to the radius through the point of contact.)

In $△\mathrm{OAP}$, applying Pythagoras's theorem
$$\begin{gathered} \Rightarrow {\left(\mathrm{OA}\right)}^2={\left(\mathrm{AP}\right)}^2+{\left(\mathrm{OP}\right)}^2 \\ \Rightarrow {13}^2=5^2+{\left(\mathrm{OP}\right)}^2 \\ \Rightarrow {\left(\mathrm{OP}\right)}^2=169-25 \\ \Rightarrow \mathrm{OP}=12 \mathrm{cm} \end{gathered}$$

Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).
 

Answer:

Statement-2 (Reason): The lengths of tangents drawn from an external point to a circle are equal.
Thus, Statement-2 is true.

Statement-1 (Assertion): A tangent to a circle is perpendicular to the radius through the point of contact.
Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Answer:

Statement-2 (Reason): The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

 

It can be observed that

OA (radius) ⊥ PA (tangent) and OB (radius) ⊥ PB (tangent)

∴ ∠OAP = 90° and ∠OBP = 90° (∵The tangent to a circle is perpendicular to the radius through the point of contact)

In quadrilateral OAPB,

Sum of all interior angles = 360º

⇒ ∠OAP +∠APB+∠PBO +∠BOA = 360º

⇒ 90º + ∠APB + 90º + ∠BOA = 360º

⇒ ∠APB + ∠BOA = 180º

So, it can be observed 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.

Answer:

Statement-2 (Reason): The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

 

It can be observed that

OA (radius) ⊥ PA (tangent) and OB (radius) ⊥ PB (tangent)

∴ ∠OAP = 90° and ∠OBP = 90° (∵The tangent to a circle is perpendicular to the radius through the point of contact)

In quadrilateral OAPB,

Sum of all interior angles = 360º

⇒ ∠OAP +∠APB+∠PBO +∠BOA = 360º

⇒ 90º + ∠APB + 90º + ∠BOA = 360º

⇒ ∠APB + ∠BOA = 180º

So, it can be observed 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.

In quadrilateral OAPB,

Sum of all interior angles = 360º

⇒ ∠OAP + ∠APB + ∠PBO + ∠BOA = 360º
⇒ ∠OAP + ∠PBO + 180º = 360º         [From (1)]
​⇒ ∠OAP + ∠PBO = 180º 
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Answer:

Statement-2 (Reason): The lengths of tangents drawn from P to the circle are equal.
Thus, Statement-2 is true.

Statement-1 (Assertion): In the given figure, PA and PB are tangents drawn from an external point P to a circle with centre O such that ∠APB = 40°. If C is a point on the circle, then ∠ACB = 70°.

Given that, ∠APB = 40°
Now,
∠APB + ∠AOB = 180°      (Angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the point of contact)

⇒ 40° + ∠AOB = 180° 
⇒ ∠AOB = 140° 
So,  ∠ACB = 70°                  (Angle subtended at the center is double the angle subtended by it at any point on the remaining part of the  circle)

Thus, Statement-2 true.

Thus, both statement are true but Statement-2 is not a correct explanation of Statement-1.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b)