SOME IMPORTANT QUESTIONS (Chapter:Circles)

Q1. ABC is an isosceles triangle in which AB = AC, circumscribed about a circle. Prove that the base is bisected by the point of contact. i.e. if AB = AC, prove that BE = EC.                 CBSE 2008

Solution:-

Since the tangents drawn from any exterior point to a circle are equal in length, therefore:-

AD = AF - ( Tangents from A)

BD = BE - ( Tangents from B)

CE = CF - ( Tangents from C)

Now,

AB = AC

AB- AD = AC- AD - (Subtracting AD from both sides)

AB- AD = AC – AF - (Since, AD=AF)

BD = CF

BE = CF

BE = CE , Hence proved.

Q2. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.

                                                                                                                               CBSE 2008,2009

Solution:-

Since the tangents drawn from any exterior point to a circle are equal in length, therefore:-

AP = AS - (Tangents from A) - (i)

BP = BQ - ( Tangents from B) - (ii)

CR = CQ - ( Tangents from C) - (iii)

DR = CQ - ( Tangents from D) - (iv)

Adding (i) , (ii) , (iii) , (iv) , we get :-

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

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

AB + CD = AD + BC

AB + CD = BC + DA , Hence proved.

Q3. Prove that a parallelogram circumscribing a circle ia a rhombus.                   CBSE 2002, 2008

Solution:-

Let ABCD be a parallelogram such that its sides touch a circle with centre O.

Since the tangents to a circle from an exterior point are equal in length , therefore:-

AP = AS - (Tangents from A) ….......(i)

BP = BQ - (Tangents from B) …........(ii).

CR = CQ - (Tangents from C) ….........(iii)

DR = DS - (Tangents from D) …...........(iv).

Adding (i) , (ii) , (iii) , (iv) , we get:-

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

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

AB + CD = AD + BC

2AB = 2BC ….....................(Since, ABCD is a parallelogram, therefore AB = CD and BC = AD)

AB = BC

AB = BC = CD = AD

Therefore, ABCD is a rhombus, Hence proved.

Q4. A circle is inscribed in a triangle ABC having sides 8cm , 10cm and 12cm. Find AD , BE and CF.                                                                                                                                 CBSE 2001

Solution:-

Since, the tangents drawn from an external point to a circle are equal , therefore:-

AD = AF = say , x

BD = BE = say , y
CE = CF = say , z

Now,

AB = 12cm , BC = 8cm and CA = 10cm

Therefore,

x + y = 12 , y + z = 8 and z + x = 10

(x+y)+(y+z)+(z+x) = 12 + 8 + 10

2(x+y+z) = 30

x+y+z = 15

Now,

x+y = 12 and x+y+z = 15

12 + z = 15

z = 15 – 12 = 3

Similarly,

y+z = 8 and x+y+z = 15

x+8 = 15

x = 15 – 8 = 7

Similarly,

z+x = 10 and x+y+z = 15

y+10 = 15

y = 15 – 10 = 5

Therefore, x = 7cm , y = 5cm ad z = 3cm i.e. AD = 7cm , BE = 5cm and CF = 3cm.

Q5. Prove that the tangents at the extremities of any chord make equal angles with the chord.

                                                                                                                             CBSE 2000,2001,2002

Solution:-

Let AB be a chord of a circle with centre O , and let AP and BP be the tangents at A and B.

Suppose the tangents meet at P. Join OP. Suppose OP meets AB at C .

We need to prove that PAC = PBC.

In triangles PCA and PCB , we have:-

PA = PB ….......(Since, tangents from an external points are equal)

  APC =  BPC ….........(Since, PA and PB are equally inclined to OP)

PC = PC ….......(Common)

therefore, by SAS criterion of congruence , we have:-

triangle PAC = triangle PBC

PAC = PBC , Hence proved.

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