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1. 10. CIRCLES

2. RECALL :
 What is a circle?
 A circle is a collection of all points in a plane which are at
a constant distance called radius and from a certain fixed
point called centre.
 A circumference which is a set of all points at a fixed
distance from the centre of the circle.
 The distance from the centre of the circle to the
circumference is called the radius of the circle.
 You can draw a circle using a compass.
Circumference

3. APPLICATIONS OF CIRCLES IN DAY TO DAY LIFE :

4. IMPORTANT TERMS RELATED TO CIRCLE :
 Chord of a circle is a line that joins two points
on the circumference of a circle.
 The diameter of a circle is its longest chord.
 An arc of a circle is a continuous part of the
circumference of the circle.
 A sector of a circle is the area/region between
an arc and the center of the circle.
 A chord divides the area of a circle into
two segments. The smaller area is
called Minor segment and the bigger area is
called Major segment.

6. Introduction to Circles
Circle and line in a plane
 For a circle and a line on a plane, there can be three possibilities.
i) If the circle and line PQ have no point in common, then we say that
PQ is a Non-intersecting line.
ii) If the circle and line PQ have only one point in common, then we
say that PQ is a Tangent to the circle.
(iii) If the circle and line PQ have two distinct points A and B, then we
say that PQ is a Secant of the circle. Also the line segment AB
is called a chord of the circle.

7. TANGENT :
A tangent to a circle is a line which touches the circle at
exactly one point.
For every point on the circle, there is a unique tangent
passing through it.
 A = Point of contact
PQ = Tangent.

8. SECANT :
SECANT :
A secant to a circle is a line which has two distinct points in
common with the circle.
It cuts the circle at two points, forming a chord of the circle.
PQ = Secant of the circle.

9. TWO PARALLEL TANGENTS AT MOST FOR A GIVEN
SECANT :
 For every given secant of a circle, there are exactly two tangents which
are parallel to it and touches the circle at two diametrically opposite
points.

10. IMPORTANT POINTS TO REMEMBER :
The number of tangents drawn from a given point.
i) If the point is in an interior region of the circle, any line
through that point will be a secant.
So, no tangent can be drawn to a circle which passes
through a point that lies inside it.
ii) When a point of tangency lies on the circle, there
is exactly one tangent to a circle that passes through it.

11. TANGENT FROM AN EXTERNAL POINT
When the point lies outside of the circle, there are accurately
two tangents to a circle through it
Tangents to a circle from an external point

12. LENGTH OF A TANGENT
The length of the tangent from the point (Say P) to the circle
is defined as the segment of the tangent from the external
point P to the point of tangency I with the circle.
 In this case, PI is the tangent length.

13. CHORD, TANGENT AND SECANT

14. CIRCLE :

15. OQ = OR + RQ = OP + RQ

16. Therefore OP ⊥ XY.
Since we know that shortest distance from a point to a line
is the perpendicular distance.
Hence proved.
This theorem is also called as TANGENT RADIUS
THEOREM.

18. R
H
S
This theorem is also called as EQUAL TANGENT LENGTHS THEOREM
O O
Q R
P P

19. IMPORTANT POINTS TO REMEMBER :
 TANGENT RADIUS THEOREM :
The tangent at any point of a circle is perpendicular to
the radius through the point of contact.
POINT OF CONTACT : The common point of a tangent to a circle
and the circle is called point of contact.
 A line drawn through the end point of the radius and perpendicular to
it is a tangent to the circle.
 EQUAL TANGENT LENGTHS THEOREM :
The length of the tangents drawn from an external point are equal.
 One and only one tangent can be drawn at any point of
a circle.

20. EX : 10.1 :
1. How many tangents can a circle have?
Answer: A circle can have infinitely many tangents since there
are infinitely many points on a circle and at each point of it,
it has a unique tangent.
2. Fill in the blanks :
(i) A tangent to a circle intersects it in ONE point(s).
(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.

21. 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line
through the centre O at a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) √119 cm
Answer :
Using the theorem, “The line drawn from the centre of the circle to the
tangent is perpendicular to the tangent”.
∴ OP ⊥ PQ .
By Pythagoras theorem in ΔOPQ,
OQ2 = OP2 + PQ2
⇒ (12)2 = 52 + PQ2
⇒PQ2 = 144 – 25 = 119
⇒PQ = √119 cm
(D) is the correct option.

22. 4. Draw a circle and two lines parallel to a given line such that
one is a tangent and the other, a secant to the circle.
Answer :
AB and XY are two parallel lines where AB is the tangent to the circle at
point C while XY is the secant to the circle.

23. EX : 10.2
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
Answer:
First, draw a perpendicular from the center O of the triangle to a point P on the circle
which is touching the tangent. This line will be perpendicular to the tangent of the circle.
So, OP is perpendicular to PQ i.e. OP ⊥ PQ

24. By using Pythagorean theorem in △OPQ,
OQ2 = OP2 + PQ2
⇒ (25)2 = OP2 + (24)2
⇒ OP2 = 625 – 576
⇒ OP2 = 49
⇒ OP = 7 cm
So, option A i.e. 7 cm is the radius of the given circle.
EX : 10.2
From the above figure, it is also seen that △OPQ is a right angled triangle.
It is given that
OQ = 25 cm and PQ = 24 cm

25. 2. In Fig. 10.11, 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°
EX : 10.2
Answer:
From the question, it is clear that OP is the radius of the circle to the tangent PT and OQ
is the radius to the tangents TQ.

26. So, OP ⊥ PT and TQ ⊥ OQ
∴ ∠OPT = ∠OQT = 90°
Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360°
So, ∠PTQ + ∠POQ + ∠OPT + ∠OQT = 360°
Now, by putting the respective values we get,
⇒ ∠PTQ + 90° + 110° + 90° = 360°
⇒ ∠PTQ = 70°
So, ∠PTQ is 70° which is option B.
EX : 10.2

27. 3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at
angle of 80°, then ∠ POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Answer:
GIVEN :
EX : 10.2
 OA is the radius to tangent PA
 OB is the radius to tangents PB.
 OA ⊥ PA and OB ⊥ PB ( Using Tangent Radius Theorem)
 So, ∠OBP = ∠OAP = 90°
 ∠A BP = 80°
TO FIND ∠POA.
In the quadrilateral AOBP,
∠AOB + ∠OAP + ∠OBP + ∠APB = 360° (Since the sum of all the interior angles will be 360°)

28. Putting their values we get,
⇒ ∠AOB + 260° = 360°
 ⇒ ∠AOB = 100°
In △OPB and △OPA,
AP = BP (Since the tangents from a point are always equal)
OA = OB (Equal radii )
OP = OP (common side)
∴ △OPB ≅ △OPA ( By SSS congruency).
∠POB = ∠POA (CPCT)
⇒ ∠AOB = ∠POA + ∠POB
⇒ 2 (∠POA) = ∠AOB
By putting the respective values we get,
⇒ ∠POA = 100°/2 = 50° ∴ ∠POA = 50°
Option A is the correct option.
EX : 10.2

29. 4. Prove that the tangents drawn at the ends of a diameter of a circle are
parallel.
Answer:
Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and OA ⊥ PQ (By tangent radius theorem)
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º
From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.
Let AB be a diameter of the circle.
Two tangents PQ and RS are drawn at points A and B respectively.
EX : 10.2

30. Q.5. Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.
Sol.
GIVEN : Let the centre of the circle is O and tangent AB touches the circle at P.
Let PX be perpendicular to AB at P.
TOPROVE : PX passes through ‘O’
PROOF : If possible let PX not passing through O.
Join OP.
Since tangent at a point to a circle is perpendicular to
the radius through that point,
∴ AB ⊥ OP i.e. ∠OPB = 90° ...(1)
But by construction,
AB ⊥ PX ⇒ ∠XPB = 90° ...(2)
From (1) and (2),
∠XPB = ∠OPB
which is possible only when O and X coincide.
Thus, the perpendicular at the point of contact to the tangent passes through the centre.
EX : 10.2

31. Q.6. The length of a tangent from a point A at distance 5 cm from the centre of the
circle is 4 cm. Find the radius of the circle.
Sol. ∵The tangent to a circle is perpendicular to the radius through the point of contact.
∠OTA = 90°
Now, in the right ΔOTA, we have:
OP
2
= OT
2
+ PT
2
⇒ 5
2
= OT
2
+ 4
2
⇒ OT
2
= 5
2
– 4
2
⇒ OT
2
= (5 – 4) (5 + 4)
⇒ OT
2
= 1 × 9 = 9 = 3
2
⇒ OT = 3
Thus, the radius of the circle is 3 cm.
EX : 10.2

32. Q.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.
Sol.
In the figure, O is the common centre, of the given concentric circles.
AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.
Since OP is the radius of the smaller circle through P, the point of contact,
∴ OP ⊥ AB (Tangent radius theorem)
⇒ ∠OPB = 90°
Also, a radius perpendicular to a chord bisects the chord.
∴ AP = BP, AB = AP + BP
Now, in right ΔAPO,
OA
2
= AP
2
– OP
2
⇒ 5
2
= AP
2
– 3
2
⇒ AP
2
= 5
2
– 3
2
⇒ AP
2
= (5 – 3) (5 + 3) = 2 × 8
⇒ AP
2
= 16 = (4)
2
⇒ AP = 4 cm
Hence, the required length of the chord AB is 8 cm.
EX : 10.2

33. I
Q.8. A quadrilateral ABCD is drawn to circumscribe a circle (see figure).
Prove that: AB + CD = AD + BC
Sol. Since the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touch the circle at P,
Q, R and S respectively, and the lengths of two tangents to a circle from an external
point are equal.
EX : 10.2
From the figure we observe that, (Using Equal tangent lengths theorem)
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)
⇒ AB + CD = AD + BC

34. 9. In Fig. 10.13, 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 :
EX : 10.2
A/q,
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)

35. Similarly,
ΔOQB ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle,
it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°

36. Given:
Consider a circle with centre O.
Let P be an external point from which two tangents PA and PB are drawn
to the circle which are touching the circle at point A and B respectively
AB is the line segment, joining point of contacts A and B together such that it subtends
∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
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:
EX : 10.2

37. Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º
∴ 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.

38. 11. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:
ABCD is a parallelogram,
∴ AB = CD ... (i) (opposite sides are equal)
∴ BC = AD ... (ii)
From the figure, we observe that,
(Using Equal tangent lengths theorem)
DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
EX : 10.2

39. Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC ... (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
⇒ 2AB = 2BC
⇒ AB = BC ... (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.

40. 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 Fig. 10.14).
Find the sides AB and AC.
Answer
In ΔABC,
Length of two tangents drawn from the same point to the circle
are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x
We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
Now semi perimeter of triangle (s) is,
⇒ 2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
⇒s = 14 + x
EX : 10.2

41. ⇒
also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)
= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]
= 2×1/2 (4x + 24 + 32) = 56 + 4x ... (ii)
Equating equation (i) and (ii) we get,
√(14 + x) 48 x = 56 + 4x
Squaring both sides,
48x (14 + x) = (56 + 4x)2
⇒ 48x = [4(14 + x)]2/(14 + x)
⇒ 48x = 16 (14 + x)
⇒ 48x = 224 + 16x
⇒ 32x = 224
⇒ x = 7 cm
Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm
Area of ΔABC = √s (s - a)(s - b)(s - c)
= √(14 + x) (14 + x - 14)(14 + x - x - 6)(14 + x - x - 8)
= √(14 + x) (x)(8)(6)
= √(14 + x) 48 x ... (i)

42. 13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend
supplementary angles at the centre of the circle.
Answer:
Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the
circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the centre of the
circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒∠1 = ∠8
EX : 10.2

43. Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary
angles at the centre of the circle.