Class 10 Maths Previous Year Questions - Circles- 2

Ans: Given: r₂ - r₁ = 7 (r₂ > r₁) ...(i)

    (From equation (i))
     ..... (ii)
Adding (i) and (ii), we get
2r₂ = 56
⇒ r₂ = 28 cm
Also, r₁ = 21 cm (From equation (i))
∴ Radius of simaller circle = 21 cm.

Ans: Given: A line T tangent to the circle at point T and O is the centre of circle.
To prove: OT ⊥ l
Construction: Take point T₁,T₂ and T₃ on line l and join OT₁, OT₂, OT₃ 
Proof: We observe that points T₁, T₂, T₃ lie outside the circle, whereas point T lies on the circle.
Hence OT₁ > OT
OT₂ > OT
OT₃ > OT
All distances OT₁, OT₂, OT₃ are greater than OT.
Only OT is the shortest distance.
Also OT = r
Hence r is the shortest distance from tangent l of the circle to the centre as we know that shortest distance between the point on line is perpendicular distance
So, OT ⊥ l

Q17: In Fig., PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents drawn at P and Q intersect at T. Find the length of TP.    [AI (C) 2017, Foreign 2015]
Ans: Given: PQ is a chord of length 8 cm
Radius OP = 5 cm. PT and QT are tangents to the circle.
To find: TP
OT is perpendicular bisector of PQ
∠ORP = 90°
(Line joining the centre of circle to the common point of two tangents drawn to circle is perpendicular bisector of line joining the point of contact of the tangents.


⇒

x² = 16
⇒ x = 3


(OR + RT)² = 25 + PT² 
(3 + y)² = 25 + PT² 
9 + y² + 6y = 25 + PT²         .......(i)
In ΔPRT, TP² = PR² + RT² 
⇒ PT² = (4)² + (y)²       .......(ii)
Put value of PT² in eq (i)
9 + y² + 6y = 25 + 16 + y² 
6y = 25 + 16 - 9
6y = 32
y = 32/6 = 16/3 cm
Putting y = 16/3 cm in eq (ii), we get

Ans: Given, ∠QPR = 120°
Radius is perpendicular to the tangent at the point of contact.
∠OQP = 90° ⇒ ∠QPO = 60°

(Tangents drawn to a circle from an external point are equally inclined to the segment, joining the centre to that point.)

Ans: PA = PB (Tangents from an external point are equal)
and ∠APB = 60°
⇒ ∠PAB = ∠PBA = 60°
∴ ΔPAB is an equilateral triangle.
Hence AB = PA = 5 cm.

Ans: Let ∠TOP = θ
∴
Hence, ∠TOS = 120°
In ∠OTS, OT = OS    (Radii of circle)
⇒

Ans: OA = 6 cm, OB = 4 cm, AP = 8 cm
OP² = OA² + AP² = 36 + 64 = 100
⇒ OP = 10 cm
BP² = OP² - OB² = 100 - 16 = 84
⇒

Ans: ∵ PA = PB ⇒ ∠BAP = ∠ABP = 50°
∴  ∠APB = 180° - 50° - 50° = 80°
and ∠AOB = 180° - 80° = 100°

Ans: ∠ACB = 90° (Angle in the semicircle)
∠CAB = 30° (given)
In ΔABC,
90° + 30° + ∠ABC = 180°
⇒ ΔABC = 60°
Now, ∠PCA = ∠ABC (Angles in the alternate segment)
∴  ∠PCA = 60°

OR
Construction: Jo in 0 to C.
∠PCO = 90° (∵ Line joining centre to point of contact is perpendicular to PQ)
In ΔAOC, OA = OC (Radii of circle)
∴ ∠OAC = ∠OCA = 30° (Equal sides have equal opp. angles)
Now, ∠PCA = ∠PCO - ∠ACO
= 90° - 30° = 60°

Ans:  Given. A circle C(0, r). PA and PB are tangents to the circle from point P, outside the circle such that ∠APB = 120°. OP is joined.

To Prove. OP = 2AP.
Construction. Join OA and OB.
Proof. Consider Δs PAO and PBO
PA = PB [Tangents to a circle, from a point outside it, are equal.]
OP = OP [Common]
∠OAP = ∠OBP = 90°

Ans: AC is tangent to circle with centre O.
Thus ∠ACO = 90°
In ΔAO'D and ΔAOC
∠ADO' = ∠ACO = 90º
∠A = ∠A  (Common)

Ans: 

(Tangents from an external point to a circle are equal)
In right ΔAET.
TA² = TE² + EA²
⇒ (12 - x)² = 64 + x²    ⇒    144 + x² - 24x = 64 + x²
⇒  x = 80/24    ⇒ x = 3.3 cm
Thus, AB = 6.6 cm

Ans: Let ABCD be a parallelogram such that its sides touch a circle with centre O.
We know that the tangents to a circle from an exterior point are equal in length.
Therefore, we have
AP = AS             (Tangents from A)      ... (i)
BP = BQ             (Tangents from B)     ... (ii)
CR = CQ              (Tangents from C)     ... (iii)
And DR = DS      (Tangents from D) ... (iv)

Adding (i), (ii), (iii) and (iv), we have
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = BC + BC      (∵ ABCD is a parallelogram ∴ AB = CD, BC = DA)
2AB = 2BC ⇒ AB = BC
Thus, AB = BC = CD = AD
Hence, ABCD is a rhombus.

Ans: Given: AP and AQ are two tangents from a pointed to a circle C (O, r).
To Prove: AP = AQ
Construction: Join OP, OQ and OA.
Proof: In order to prove that AP = AQ, we shall first prove that ΔOPA ≅ ΔOQA.
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OP ⊥ AP and OQ ⊥ AQ
⇒ ∠OPA = ∠OQA = 90°    ........(i)

Now, in right triangles OPA and OQA, we have
OP = OQ    (Radii of a circle)
∠OPA = ∠OQA    (Each 90°)
and OA = OA    (Common)
So, by RHS-criterion of congruence, we get

Hence, lengths of two tangents from an external point are equal.