10. CIRCLES (NCERT)
NCERT
10. CIRCLES
EXERCISE 10.1
1. How many tangents can a circle have ?
Solution: A circle have infinitely many tangents .
2. Fill in the blanks :
(i) A tangent to a circle intersects it in 
points .
(ii) A line intersecting a circle in two points is called a 
.
(iii) A circle can have parallel tangents at the most .
(iv) The common point of a tangent to a circle and the circle is called .
Solution: (i) A tangent to a circle intersects it in 
points .
(ii) A line intersecting a circle in two points is called a 
.
(iii) A circle can have 
parallel tangents at the most .
(iv) The common point of a tangent to a circle and the circle is called 
.
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 
. Length PQ is :
(A) 12 cm (B) 13 cm
(C) 8.5 cm (D) 
cm
Solution: In 
, we have
(D) cm .
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 .
Solution: In Figure , O be a centre of the circle.
EF is given line , CD is a tangent and AB is a secant of the circle . So, line EF parallel to secant AB and also the line EF parallel to tangent CD .
EXERCISE 10.2
1.From a point Q, the length of the tangent to a circle is 24cm and the distance of Q from the centre is 25cm .The radius of the circle is
(A) 7cm (B) 12cm (C) 15cm (D) 24.5cm
Solution : Here , OQ = 25 cm and QT = 24 cm
In 
OQT , we have
cm
Answer: (A) 7 cm
2. In fig. 10.11,if TP and TQ are the two tangents to a circle with centre O so that 
then 
is equal to :
(A) 60° (B) 70° (C) 80° (D) 90°
Solution: Here, 
and ∠OQT=90°
Since, OPTQ is cyclic quadrilateral , We have
Answer : (B) 70°
3. If the tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then 
is equal to
(A) 50° (B) 60° (C) 70° (D) 80°
Solution: Since , 
and 
In 
, we have
Answer : (A) 50°
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre .
Solution:
6. The length of a tangent from a point A at distance 5cm from the centre of the circle is 4cm.Find the radius of the circle .
Solution: In 
, We have
cm
7. Two concentric circles are of radii 5cm and 3cm .Find the length of the chord of the larger circle which touches the smaller circle.
Solution: Here, OM = 3 cm and OP = 5 cm .
In OMP, we have
cm
Since OM
PQ
cm
8. A quadrilateral ABCD is drawn to circumscribe a circle a circle (see Fig. 10.12). Prove that 
Solution: let, 
be a quadrilateral and its sides are 
, 
,
and 
in which the points 
and 
are touch of a circle with centre O respectively .
To prove : 
.
Proof : Since the length of the two tangents from an external point to a circle are equal .
(From 
)…….(i)
(From 
)…….(ii)
(From 
)…….(iii)
(From D )…….(iv)
(i)
(ii)(iii)(iv) 
Proved.
9. In Fig. 10.13, XY and 
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 
.
Solution: Given , 
and ' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting at A and ' at B . To Prove
Construction : Join 
and
.
Proof : Since the lengths of the two tangents from an external point to a circle are equal and also the subtend equal angles at the centre .
Therefore , 
is a bisector of 
and 
is a bisect of 
Since , 
and is a transversal
[From 
and 
] 
we have,
[ From 
]
Proved.
10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the nagle subtended by the line –segment joining the points of contact at the centre.
Solution: Given, AP and BP be two tangents drawn from an external point P to a circle with centre O .
To prove : 
Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact .
Therefore , 
and 
.
So, 
and 
.
OAPB is cyclic quadrilateral , we have
proved .
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Solution: Given , be a parallelogram and its sides ,
,
and
are touch the circle at 
, 
, 
and 
respectively .
To prove : is a rhombus .
Proof : Since the lengths of tangents drawn from an external point to a circle are equal .
Therefore , (From ) ……. 
(From B )…….
(From C )…….
(From D )……. 
Since, be a parallelogram . Thus, 
and 
So , 
Therefore, is a rhombus.
12. A triangle ABC is drawn circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
Solution: Since, the lengths of the two tangents from an external point to a circle are equal.
cm ; 
cm ; let, 
cm
Therefore, 
cm , 
cm , 
cm
; 
; 
So , 
cm
Ar
Ar 
+ Ar 
+ Ar
A/Q, 
(impossible)
or 
Thus, 
cm and 
cm
13. Prove that the opposite sides of a quadrilateral circumscribing a circle subtended supplementary angles at the centre of the circle.
Solution: Given , ABCD is a quadrilateral circumscribing a circle with centre O and touches the quadrilateral at P , Q , R and S respectively .
To prove : (i) 
(ii) 
Construction : Join OP , OQ , OR and OS respectively .
Proof : Since the lengths of the two tangents from an external point to a circle are equal .
i.e., 
; 
; 
and 
.
In AOS and AOP , we have
(Given)
[ Common side]
[ R.H.S rule]
[ C.P.C.T] 
Similarly , 
; 
; 
Let, 
,
, 
, 
, 
, 
, 
and 
We know that the sum of the all angles of subtended at a point is 360° .
Similarly , Proved.
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