NCERT
10th Mathematics
Exercise 10.2
Topic: Circle
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In Q.1 to 3, choose the correct option and give justification.
1. From a point Q, the length of the tangent to a circle is 24 cm, and the
distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm
Solution:
1) QT is the tangent to the circle with center O at point T and QT = 24 cm.
2) OQ = 25 cm.
3) Let the radius OT be x cm.
4) In ∆ OTQ, by the theorem of Pythagoras, we get,
(OT)2 + (QT)2 = (OQ)2
(OT)2 = (OQ)2 – (QT)2
(OT)2 = (25)2 – (24)2
(OT)2 = (25 – 24) (25 + 24)
(OT)2 = 1(49)
(OT)2 = 49
OT = √49
OT = 7
5) Therefore, answer is (A), OT = 7 cm.
2. In the following fig., if TP and TQ are the two tangents to a circle
with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to
(A) 60° (B) 70° (C) 80° (D) 90°
Solution:
1) TQ and TP are the tangents to the circle with center O at points Q and P.
2) ∠ POQ = 110°.
3) In □ OPTQ,
∠ OQT = 90°
∠ OPT = 90°∠ PTQ + ∠ POQ = 180°
∠ PTQ + 110° = 180°
∠ PTQ = 180° – 110°
∠ PTQ = 70°
4) Therefore, answer is (A), ∠ PTQ = 70°.
3. If tangents PA and PB from point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠ POA is equal to
(A) 50° (B) 60° (C) 70° (D) 80°
Solution:
1) PA and PB are the tangents to the circle with center O at points A and B.
2) ∠ APB = 80°.
3) In ∆ POA and ∆ POB,
∠ PAO = ∠ PBO = 90°.
OA = OB radii of the same circle.
OP = OP common side of ∆ POA and ∆ POB.
∆ POA ≅ ∆ POB by Hypotenuse-side theorem -------- 1
4) From 1, using CACT, we have,
∠ OPA = ∠ OPB ------------- 2
∠ OPA + ∠ OPB = 80° ------------- 3 Given
5) From 2 and 3, we have,
∠ OPA + ∠ OPB = 80°
∠ OPA + ∠ OPBA = 80°
2 ∠ OPA = 80°
∠ OPA = 40° ------------- 4
6) In ∆ POA and From 4, we have,
∠ POA + ∠ OPA = 90°
∠ POA + 40° = 90°
∠ POA = 90° – 40°
∠ POA = 50°
7) Therefore, answer is (A), ∠ POA = 50°.
4. Prove that the tangents drawn at the ends of the diameter of a circle are
parallel.
1) AB and CD are the tangents to the circle with center O at points P and Q.
2) OQ ⊥ tangent CD and OP ⊥ tangent AB.
3) PQ is the diameter of a circle with center O.
4) So,
∠ OQC = 90°
∠ OPB = 90°
5) Here,line PQ is the transversal on the line AB and line CD, so ∠ OQC
and ∠ OPB are alternate interior angles.
6) As, alternate interior angles ∠ OQC = ∠ OPB, line AB and line CD are
parallel. Hence proved.
5. Prove that the perpendicular at the point of contact to the tangent to a
circle passes through the centre.
1) A tangent AB touching the circle at point P.
2) We know that the tangent of a circle is ⊥ to radius at point of contact,
therefore,
OP ⊥ tangent AB.
∠ OPA = 90° ----------------- equation 1
3) Now let us cosider that, QP ⊥ AB, so we have,
∠ QPA = 90° ----------------- equation 2
4) From equations 1 and 2, we can say that.
∠ OPA = ∠ QPA = 90°, is possible only if line QP passes through O.
5) So, the perpendicular at the point of contact to the tangent to a
circle passes through the centre is proved.
6. The length of a tangent from point A at a distance of 5 cm from the centre
of the circle is 4 cm. Find the radius of the circle.
Solution:
1) QT is the tangent to the circle with center O at point T and QT = 4 cm.
2) OQ = 5 cm.
3) Let the radius OT be x cm.
4) In ∆ OTQ, by the theorem of Pythagoras, we get,5) Radius of a circle OT = 3 cm.
7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the
chord of the larger circle which touches the smaller circle.
Solution:
1) The chord AB of larger circle touches the smaller circle at P.
2) OP = 3 cm and OB = 5 cm.
3) In ∆ OPB, by the theorem of Pythagoras, we get,4) Length of the chord of the larger circle AB = 2 x PB = 2 x 4 = 8 cm.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see following fig.).
Prove that AB + CD = AD + BC.
1) AB, BC, CD, and DA are tangents to the circle touching at points P, Q, R,
and S respectively.
2) As, the lengths of tangents drawn from an external point to a circle are
equal, we have,
AP = AS ------------- equation 1
BP = BQ ------------- equation 2
CR = CQ ------------- equation 3
DR = DS ------------- equation 4
3) Adding equations 1, 2, 3, and 4, we get,
AP + BP + CR + DR = AS + BQ + CQ + DS ------------- equation 5
4) Rearranging the terms of equation 5, we get,
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
(AB) + (CD) = (AD) + (BC)
5) So, AB + CD = AD + BC is proved.
9. In the following fig., 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. Prove that ∠ AOB = 90°.
Solution:
1) XY and X’Y’ are two parallel tangents to a circle with centre O, touching at
points P and Q respectively.
2) AB is another tangent at C to the circle and intersects XY at A and X'Y' at B.
a) Tangents are ⊥ to the radii of a circle.
∠ OPA = ∠ OCA = 90° --------------- equation 1
b) As, the lengths of tangents drawn from an external point to a circle are
equal, we have,
AP = AC --------------- equation 2
c) Common side
AO = AO --------------- equation 3
4) From equation 1, 2, 3, and hypotenuse-side theorem we have,
∆ AOP ≅ ∆ AOC
5) So, using CPCT we have,
∠ POA = ∠ COA --------------- equation 4
∠ POC = ∠ POA + ∠ COA --------------- equation 5
6) From equation 4, and 5, we have
∠ POC = ∠ POA + ∠ COA
∠ POC = ∠ COA + ∠ COA
∠ POC = 2 ∠ COA --------------- equation 6
7) In the same way, we can prove that,
∠ QOC = 2 ∠ COB --------------- equation 7
8) Adding equations 6, and 7, we get,
2 ∠ COA + 2 ∠ COB = ∠ POC + ∠ QOC
2 ∠ COA + 2 ∠ COB = 180°
2 (∠ AOB) = 180°
∠ AOB = 90°, hence proved.
10. Prove 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.
Solution:
1) TQ and TP are the tangents to the circle with center O at points Q and P.
2) ∠ POQ = 110°.
3) In □ OPTQ, as tangents are ⊥ to the radii of a circle,
∠ OQT = 90° ------------- equation 1
∠ OPT = 90° ------------- equation 2
4) In □ OPTQ and from equations 1, and 2,
∠ PTQ + ∠ POQ + ∠ OQT + ∠ OPT = 360°
∠ PTQ + ∠ POQ + 90° + 90° = 360°
∠ PTQ + ∠ POQ + 180° = 360°∠ PTQ + ∠ POQ = 180°
5) ∠ PTQ and ∠ POQ are supplimentary.
11. Prove that the parallelogram circumscribing a circle is a rhombus.
1) AB, BC, CD, and DA are tangents to the circle touching at points P, Q, R,
and S respectively.
2) As, the lengths of tangents drawn from an external point to a circle are
equal, we have,
AP = AS ------------- equation 1
BP = BQ ------------- equation 2
CR = CQ ------------- equation 3
DR = DS ------------- equation 4
3) Adding equations 1, 2, 3, and 4, we get,
AP + BP + CR + DR = AS + BQ + CQ + DS ------------- equation 5
4) Rearranging the terms of equation 5, we get,
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
(AB) + (CD) = (AD) + (BC) ------------- equation 6
5) As, ABCD is parallelogram, we have,
CD = AB ------------- equation 7
BC = AD ------------- equation 8
6) From equations 6, 7, and 8, we have
(AB) + (CD) = (AD) + (BC)
(AB) + (AB) = (AD) + (AD)
2(AB) = 2(AD)
AB = AD ------------- equation 9
7) From equations 7, 8, and 9, we can say that all the sides of the
parallelogram are equal, so ABCD is the rhombus. Hence proved.
12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that
the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see the following fig.). Find the sides AB and AC.
1) AB, BC, and CA are tangents to the circle touching at points E, D, F.
2) As, the lengths of tangents drawn from an external point to a circle are
equal, we have,
CD = CF = 6
BD = BE = 8
AE = AF = x
3) So, we have,
a = BC = 8 + 6 = 14 ------------- equation 1
b = CA = (x + 6) ------------- equation 2
c = AB = (x + 8) ------------- equation 3
4) So, we have,
s = (a + b + c)/2
s = (14 + x + 6 + x + 8)/2
s = (2x + 14 + 6 + 8)/2
s = (x + 7 + 3 + 4)
s = (x + 14)
5) Using Heron's formula, we have,
A△(ABC) = √[s(s - a)(s - b)(s - c)]
A△(ABC) = √[(x + 14)(x + 14 - 14)(x + 14 - x - 6)(x + 14 - x - 8)]
A△(ABC) = √[(x + 14)(x)(8)(6)]
A△(ABC) = √[48x(x + 14)] ------------- equation 4
6) We know that,
A△(ABC) = A△(AOB) + A△(BOC) + A△(COA)
A△(ABC) = [(1/2)x(4)x(c)] + [(1/2)x(14)x(a)] + [(1/2)x(4)x(b)]
A△(ABC) = [(1/2)x(4)x(x + 8)] + [(1/2)x(4)x(14)] + [(1/2)x(4)x(x + 6)]
A△(ABC) = [(2)x(x + 8)] + [(2)x(14)] + [(2)x(x + 6)]
A△(ABC) = 2[(x + 8) + (14) + (x + 6)]
A△(ABC) = 2[2x + 8 + 14 + 6]
A△(ABC) = 4[x + 4 + 7 + 3]
A△(ABC) = 4[x + 14] ------------- equation 5
7) From equation 1, and 2, we have,
√[48x(x + 14)] = 4[x + 14]
[48x(x + 14)] = 16[x + 14]2
x = 7 ------------- equation 6
8) Put x = 7 from equation 6 in euations 2 and 3, we have,
b = CA = 13 ------------- equation 7
c = AB = 15 ------------- equation 8
CA = 13 cm and AB = 15 cm.
13. Prove that opposite sides of a quadrilateral circumscribing a circle
subtend supplementary angles at the centre of the circle.
1) AB, BC, CD, and DA are tangents to the circle touching at points P, Q, R,
and S respectively.
2) ∆ AOP and ∆ AOS, we have,
AP = AS ------------- tangent segments are same,
OA = OA ------------- common side of two triangles,
OP = OS ------------- radii of the same circle.
3) Using sss theorem, we have,
∆ AOP ≅ ∆ AOS ------------- by CPCT are equal, we have,
∠ a = ∠ h ------------- equation 1
4) Simillarly, we get,
∠ g = ∠ f ------------- equation 2
∠ e = ∠ d ------------- equation 3
∠ c = ∠ b ------------- equation 4
5) Adding all central angles, we get
∠ a + ∠ g + ∠ e + ∠ c + ∠ h + ∠ f + ∠ d + ∠ b = 360° ------------- equation 5
6) Rearranging the terms of equation 5, we get,
(∠ a + ∠ h) + (∠ g + ∠ f) + (∠ e + ∠ d) + (∠ c + ∠ b) = 360° ----- equation 6
7) Put ∠ h = ∠ a, ∠ g = ∠ f, ∠ d = ∠ e, ∠ c = ∠ b from 1, 2, 3, and 4, in 6,
(∠ a + ∠ a) + (∠ f + ∠ f) + (∠ e + ∠ e) + (∠ b + ∠ b) = 360°
2 ∠ a + 2 ∠ f + 2 ∠ e + 2 ∠ b = 360°
2 (∠ a + ∠ f + ∠ e + ∠ b) = 360°
(∠ a + ∠ f + ∠ e + ∠ b) = 180°
(∠ a + ∠ b) + (∠ e + ∠ f) = 180°
(∠ AOP + ∠ BOP) + (∠ COR + ∠ DOR) = 180°
(∠ AOB) + (∠ COD) = 180° ------------- equation 7
8) Simillarly, we can prove that,
(∠ AOD) + (∠ BOC) = 180° ------------- equation 8
9) From equations 7, and 8, we can say that,
opposite sides of a quadrilateral subtend supplementary angles at the centre of the circle. Hence proved.
