NCERT Grade 10-Circles-Answers

NCERT Solutions for Class 10 Maths

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1.     A circle can have infinitely many tangents since there are infinitely many points on the circumference of the circle and at each point of it, it has a unique tangent.

2.    (i) A tangent to a circle intersects it in one point.

(ii) A line intersecting a circle in two points is called a secant.

(iii) A circle can have two parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called point of contact.

3.     (D) ∵  PQ is the tangent and OP is the radius through the point of contact.

∴ ∠ OPQ = 900              [The tangent at any point of a circle is ⊥ to the radius  through the point of contact]

∴ In right triangle OPQ, O

Q2 = OP2 + PQ2             [By Pythagoras theorem]

⇒ (12)2 = (5)2 + PQ2        ⇒        144 = 25 + PQ2

⇒ PQ2 = 144 – 25 = 119

⇒ PQ = √119 cm

4.    

5.    (A) ∵ ∠ OPQ = 900

[The tangent at any point of a circle is ⊥ to the radius through the point of contact]

∴ In right triangle OPQ,

OQ2 = OP2 + PQ2 [By Pythagoras theorem]

⇒ (25)= OP2 +(24)625           ⇒         OP2 + 576

⇒ OP2 = 625 – 576 = 49               ⇒         OP = 7 cm

(B) ∠ POQ = 1100, ∠ OPT = 900 and ∠ OQT = 900

[The tangent at any point of a circle is ⊥ to the radius through the point of contact] In quadrilateral OPTQ,

∠ POQ + ∠ OPT + ∠ OQT + ∠ PTQ = 3600

[Angle sum property of quadrilateral]

⇒ 1100 + 900 + 900 + ∠ PTQ = 3600

⇒ 2900 + ∠ PTQ = 3600         ⇒      ∠ PTQ = 700

6.   (A) ∵ ∠ OPQ = 900

[The tangent at any point of a circle is ⊥ to the radius through the point of contact]

∠ OPA = 2 –  BPA    [Centre lies on the bisector of the angle between the two tangents]

In Δ OPA,

∠ OAP + ∠ OPA + ∠ POA = 1800      [Angle sum property of a triangle]

⇒ 900 + 400 + ∠ POA = 1800

⇒ 1300 + ∠ POA = 1800           ⇒          ∠ POA = 500

7.   Given: PQ is a diameter of a circle with centre 0. The lines AB and CD are the tangents at P and Q respectively.

To Prove: AB || CD

Proof: Since AB is a tangent to the circle at P and OP is the radius through the point of contact.

∴ ∠ OPA = 900               … (i)

[The tangent at any point of a circle is ⊥ to the radius through the point of contact]

∵ CD is a tangent to the circle at Q and OQ is the radius through the point of contact.

∴ ∠ OQD = 900           … (ii)

[The tangent at any point of a circle is ⊥ to the radius through the point of contact]

From eq. (i) and (ii), ∠ OPA = ∠ OQD

But these form a pair of equal alternate angles also,

∴ AB || CD

8.   We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact and the radius essentially passes through the centre of the circle, therefore the perpendicular at the point of contact to the tangent to a circle passes through the centre.

9.    We know that the tangent at any point of a circle is ⊥ to the radius through the point of contact.

∴ ∠ OPA = 900

∴  OA2 = OP2 + AP2         [By Pythagoras theorem]

⇒ (5)2 = (OP)+ (4)2

⇒ 25 = (OP)2 +16          ⇒          OP2 = 9

⇒ OP = 3 cm

10.  Let 0 be the common centre of the two concentric circles. Let AB be a chord of the larger circle which touches the smaller circle at P.

Join OP and OA.

Then, ∠ OPA = 900

[The tangent at any point of a circle is ⊥ to the radius through the point of contact

∴ 0A2 = OP2 + AP2            [By Pythagoras theorem]

⇒ (5)2 = (3)+ AP2

⇒ 25 = 9 + AP2                ⇒            AP2 = 16

⇒ AP = 4 cm

Since the perpendicular from the centre of a circle to a chord bisects the chord, therefore

AP = BP = 4 cm

⇒ AB = AP + BP = AP + AP = 2AP = 2 x 4 = 8 cm

11.   We know that the tangents from an external point to a circle are equal.

∴ AP = AS                          … (i)

   BP = BQ                         … (ii)

   CR = CQ                         … (iii)

   DR = DS                         … (iv)

On adding eq. (i), (ii), (iii) and (iv), we get

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

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

12.   Given: In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

To Prove: ∠ AOB = 90°

Construction: Join OC

Proof:    OPA = 900            (i)

 ∠ OCA = 900            (ii)

[Tangent at any point of a circle is ⊥ to the radius through the point of contact]

In right angled triangles OPA and OCA,

OA = OA           [Common]

AP = AC           [Tangents from an external point to a circle are equal]

∴ Δ OPA  ≅ Δ OCA                     [RHS congruence criterion]

∴ ∠ OAP = ∠ OAC          [By C.P.C.T.]

Similarly,                   ∠ OBQ = ∠ OBC

∵ XY || X’Y’ and a transversal AB intersects them.

∴ ∠ PAB + ∠ QBA = 1800        [Sum of the consecutive interior angles on the same side of the transversal is 1800]

⇒ ∠ OAC + ∠ OBC = 900                         [From eq. (iii) & (iv)]

In Δ AOB,

∠ OAC + ∠ OBC + ∠ AOB = 1800       [Angel sum property of a triangle]

⇒ 900 + ∠ AOB = 1800                            [From eq. (v)]

⇒ ∠ AOB = 900

13.   ∠ OPA = 900              …(i)

  OCA = 900             …(ii)

[Tangent at any point of a circle is ⊥ to the radius through the point of contact]

∵ OAPB is quadrilateral.

∴ ∠ APB + ∠ AOB + ∠ OAP + ∠ OBP = 3600

[Angle sum property of a quadrilateral]

⇒ ∠ APB + ∠ AOB + 900 + 900 = 3600

[From eq. (i) & (ii)]

⇒ ∠ APB + ∠ AOB = 1800

∴ ∠ APB and ∠ AOB are supplementary.

14.  Given: ABCD is a parallelogram circumscribing a circle.

To Prove: ABCD is a rhombus.

Proof: Since, the tangents from an external point to a circle are equal.

∴     AP = AS                         … (i)

BP = BQ                                … (ii)

CR = CQ                               … (iii)

DR = DS                               … (iv)

On adding eq. (i), (ii), (iii) and (iv), we get

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

⇒ AB + CD = (AS + DS) + (BQ + CQ) AB + CD = AD + BC

⇒ AB + AB = AD + AD                 [Opposite sides of || gm are equal]

⇒ 2AB = 2AD

⇒ AB = AD

But           AB = CD        and AD = BC        [Opposite sides of || gm]

∴ AB = BC = CD = AD

∴ Parallelogram ABCD is a rhombus.

15.  Join OE and OF. Also join OA, OB and OC.

Since BD = 8 cm

∴ BE = 8 cm

[Tangents from an external point to a circle are equal]

Since CD = 6 cm

∴ CF = 6 cm

[Tangents from an external point to a circle are equal]

Let           AE = AF = x

Since       OD = OE = OF = 4 cm

[Radii of a circle are equal]

Now, Area of Δ ABC = Area of Δ OBC + Area of Δ OCA + Area of Δ OAB

Squaring both sides,

(x + 14) (x) (6) (8) = 16 (x + 14)2       ⇒           3x = x + 14

⇒ 2x = 14            ⇒           x = 7

∴           AB = x+8 = 7 + 8 = 15 cm

And      AC = x+ 6 = 7 + 6 = 13 cm

16.  Given: ABCD is a quadrilateral circumscribing a circle whose centre is 0.

To Prove: (i) ∠ AOB + ∠ COD = 1800 (ii) ∠ BOC + ∠ AOD = 1800

Construction: Join OP, OQ, OR and OS.

Proof: Since tangents from an external point to a circle are equal.

∴ AP = AS,

BP = BQ ……………… (i)

CQ = CR

DR = DS

In Δ OBP and Δ OBQ,

OP = OQ                      [Radii of the same circle]

OB = OB                      [Common]

BP = BQ                       [From eq. (i)]

∴ Δ OPB ≅ Δ OBQ      [By SSS congruence criterion]

∴ ∠ 1 = ∠ 2                   [By C.P.C.T.]

Similarly,             ∠ 3 = ∠4, ∠ 5 = ∠ 6, ∠ 7 = ∠ 8

Since, the sum of all the angles round a point is equal to 3600.

∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 3600

∠1 + ∠1 + ∠4 + ∠4 + ∠5 + ∠5 + ∠8 + ∠8 = 360°

2 (∠1 + ∠4 + ∠5 + ∠8) = 3600

∠1 + ∠4 + ∠5 + ∠8 = 1800 (∠1 + ∠5) + (∠4 + ∠8) = 1800

∠ AOB + ∠ COD = 1800

Similarly we can prove that

∠ BOC + ∠ AOD = 1800.

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