ROUTERA


Chapter 8 Circles

Class 10th R. S. Aggarwal Maths Solution
CBSE Class 10 Maths
R. S. Aggarwal Solution


Circles Exercise Ex. 8A

Solution 1

  

Let O be the centre of the circle and PT be the length of tangent from P to the circle.

Then, OT = 20 cm and OP = 29 cm

Since tangent to a circle is always perpendicular to the radius through the point of contact.

∠OTP = 90o

In right ΔOTP,

OP2 = OT2 + PT2

292 = 202 + PT2

PT2 = 292 - 202 = (29 - 20)(29 + 20) = 9 × 49 = 441

PT = 21 cm

Hence, the length of the tangent drawn from P to the circle is 21 cm.

Solution 2

PA is the tangent to the circle with centre O and radius, such that PO = 25 cm, PA = 24 cm

In PAO, A = 90,

By Pythagoras theorem:

Hence, the radius of the circle is 7 cm.

Solution 3

  

  


Solution 4

Solution 5

Given AP is a tangent at A and OA is radius through A and PA and PB are the tangent segments to circle with centre O.

Therefore, OA is perpendicular to AP, similarly, OB is perpendicular to BP.


OAP = 90

And OBP = 90

So, OAP = OBP = 90

OBP + OAP = (90 + 90) = 180

Thus, the sum of opposite angles of quad. AOBP is 180

AOBP is a cyclic quadrilateral

Solution 6

  

  

Solution 7

Given: From an external point P, tangent PA and PB are drawn to a circle with centre O. CD is the tangent to the circle at a point E and PA = 14cm.

      

Since the tangents from an external point are equal, we have

PA = PB,

Also, CA = CE and DB = DE

Perimeter of PCD = PC + CD + PD

=(PA - CA) + (CE + DE) +(PB - DB)

= (PA - CE) + (CE + DE) + (PB - DE)

= (PA + PB) = 2PA = 2(14) cm

= 28 cm

Hence, the Perimeter of PCD = 28 cm

Solution 8

A circle is inscribed in a triangle ABC touching AB, BC and CA at P, Q and R respectively.

Also, AB = 10 cm, AR = 7cm, CR = 5cm

AR, AP are the tangents to the circle

AP = AR = 7cm

AB = 10 cm

BP = AB - AP = (10 - 7)= 3 cm

Also, BP and BQ are tangents to the circle

BP = BQ = 3 cm

Further, CQ and CR are tangents to the circle

CQ = CR = 5cm

BC = BQ + CQ = (3 + 5) cm = 8 cm

Hence, BC = 8 cm

Solution 9

Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively

We know that the length of tangents drawn from an exterior point to a circle are equal

AP = AS ----(1)    {tangents from A}

BP = BQ ---(2)     {tangents from B}

CR = CQ ---(3)    {tangents from C}

DR = DS----(4)    {tangents from D}

Adding (1), (2) and (3) we get

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AD = (AB + CD) - BC = {(6 + 4) - 7} cm = 3 cm

Hence, AD = 3 cm

Solution 10

Solution 11

Given O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. PA = 10cm. Join OA, OB and OP.

Then, OB = 4 cm, OA= 6 cm and PA = 10 cm

In triangle OAP,

Hence, BP = 10.9 cm

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Circles Exercise Ex. 8B

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Solution 15

 

  

PM and PN are tangents to a circle.

OM ⏊ PM and ON ⏊ PN

∠OMP = 90o and ∠ONP = 90o

In ΔOMP and ΔONP,

OM = ON  (radii of same circle)

∠OMP = ∠ONP = 90o 

OP = OP  (common)

ΔOMP ≌ ΔONP (RHS congruency)

∠OPM = ∠OPN  (CACT)

∠OPM

Hence, the length of OP is  .  

Circles Exercise MCQ

Solution 1

Correct option : (b)

  

We can draw only 2 tangents from an external point to a circle.

Solution 2

Solution 3

  

Solution 4

Correct option: (d)

The 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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Solution 50

Correct option: (d)

Options (a), (b) and (c) are all true.

However, option (d) is false since we can draw only parallel tangents on either side of the diameter, which would be parallel to a given line.

Solution 51

Correct option: (d)

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.

  

Solution 52

Correct option: (d)

Options (a), (b) and (c) are 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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Solution 55

  

Circles Exercise Test Yourself

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Solution 9

  1. A line intersecting a circle in two distinct points is called a secant.
  2. A circle can have two parallel tangents at the most.
  3. The common point of a tangent to a circle and the circle is called the point of contact.
  4. A circle can have infinitely many tangents.

 

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