Q.6: I
n the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then find the length of each tangent.
Solution:
Construction: Join AC and BC.
Proof: ∠1 = ∠2 = 90° ….[Tangent is I to the radius (through the point of contact
∴ APBC is a square.
Length of each tangent
= AP = PB = 4 cm
= AC = radius = 4 cm
Q.7: In the given figure, PQ and PR are two tangents to a circle with Centre O. If ∠QPR = 46°, then calculate ∠QOR.
Solution:
∠OQP = 900 ∠ORP = 90° ∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° …[Angle sum property of a quad. 90° + 46° + 90° + ∠QOR = 360° ∠QOR = 360° – 90° – 46° – 90° = 134°
Q.8:
From an external point P, tangents PA and PB are drawn to a circle with centre 0. If ∠PAB = 50°, then find ∠AOB.
Solution:
PA = PB …[∵ Tangents drawn from external point are equal
∠PBA = ∠PAB = 50° …[Angles equal to opposite sides
In ∆ABP, ∠PBA + ∠PAB + ∠APB = 180° …[Angle-sum-property of a ∆
50° + 50° + ∠APB = 180°
∠APB = 180° – 50° – 50° = 80°
In cyclic quadrilateral OAPB
∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180°
∠AOB + 80o = 180°
∠AOB = 180° – 80° = 100°
∠PBA = ∠PAB = 50° …[Angles equal to opposite sides
In ∆ABP, ∠PBA + ∠PAB + ∠APB = 180° …[Angle-sum-property of a ∆
50° + 50° + ∠APB = 180°
∠APB = 180° – 50° – 50° = 80°
In cyclic quadrilateral OAPB
∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180°
∠AOB + 80o = 180°
∠AOB = 180° – 80° = 100°
Q.9:
If two tangents inclined at an angle
60
∘
are drawn to a circle of radius
3
c
m
,
then the length of each tangent is
(A)
3
2
√
3
c
m
(B)
6
c
m
(C)
3
c
m
(D) 3
√
3
c
m
Solution:
Let
P
be an external point and a pair of tangents is drawn from point P and angle between these two tangents is
60
∘
Radius of the circle
=
3
c
m
Join OA and OP
Also, OP is a bisector line of
∠
APC
∴
∠
A
P
O
=
∠
C
P
O
=
30
∘
O
A
⊥
A
P
Also, tangents at any point of a circle is perpendicular to the radius through the point of contact.
In right angled
Δ
O
A
P
,
we have
tan
30
∘
=
O
A
A
P
=
3
A
P
⇒
1
√
3
=
3
A
P
⇒
A
P
=
3
√
3
c
m
A
P
=
C
P
=
3
√
3
c
m
[Tangents drawn from an external point are equal]
Hence, the length of each tangent is
3
√
3
c
m
.
Radius of the circle
=
3
c
m
Join OA and OP
Also, OP is a bisector line of
∠
APC
∴
∠
A
P
O
=
∠
C
P
O
=
30
∘
O
A
⊥
A
P
Also, tangents at any point of a circle is perpendicular to the radius through the point of contact.
In right angled
Δ
O
A
P
,
we have
tan
30
∘
=
O
A
A
P
=
3
A
P
⇒
1
√
3
=
3
A
P
⇒
A
P
=
3
√
3
c
m
A
P
=
C
P
=
3
√
3
c
m
[Tangents drawn from an external point are equal]
Hence, the length of each tangent is
3
√
3
c
m
.
Q.10: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Solution:
Given:
A
circle with center
O
with tangent
X
Y
at point of contact
P
.
To Prove:
O
P
⊥
X
Y
Proof: Let
Q
be a point on
X
Y
connect
O
Q
Suppose it touches the circle at
R
Hence,
O
Q
>
O
R
O
Q
>
O
P
(
∵
O
P
=
O
R
)
(radius)
Same will be the case with all other points on the circle
Hence,
We get
O
P
is the smallest line that connects
X
Y
.
Q.11.
Two concentric circles are of radii 7 cm and r cm respectively, where r > 7. A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of r.
Solution:
Given: OC = 7 cm, AB = 48 cm
To find: r = ?
∠OCA = 90° ..[Tangent is ⊥ to the radius through the point of contact
∴ OC ⊥ AB
AC =
1
2
(AB) … [⊥ from the centre bisects the chord
⇒ AC =
1
2
(48) = 24 cm
In rt. ∆OCA, OA
2
= OC
2
+ AC
2
… [Pythagoras’ theorem
r
2
= (7)
2
+ (24)
2
= 49 + 576 = 625
∴ r=
625
−−−√
= 25 cm
Q.12:
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
ABCD is a ॥
gm
.
To prove. ABCD is a rhombus.
Proof. In ॥
gm
, opposite sides are equal
AB = CD
and AD = BC ..(i)
AP = AS …[Tangents drawn from an external point are equal in length
PB = BQ
CR = CO
DR = DS
By adding these tangents,
(AP + PB) + (CR + DR) = AS + BQ + CQ + DS
AB + CD = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = BC + BC … [From (i)
2AB = 2 BC
AB = BC …(ii)
From (i) and (ii), AB = BC = CD = DA
∴ ॥
gm
ABCD is a rhombus.
Q. 13:
In the given figure, an isosceles ∆ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
Solution:
The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F F respectively.
AB = AC
To prove: BD = CD
Proof: Since the lengths of tangents drawn from an external point to a circle are equal
∴ AF = AE … (i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC … [Given
∴ CD = BD
AB = AC
To prove: BD = CD
Proof: Since the lengths of tangents drawn from an external point to a circle are equal
∴ AF = AE … (i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC … [Given
∴ CD = BD
Q.14:
In the given figure, a circle inscribed in ∆ABC touches its sides AB, BC and AC at points D, E & F K respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.
Solution:
Let AD = AF = x
BD = BE = y …[Two tangents drawn from and an external point are equal
CE = CF = z
AB = 12 cm …[Given
∴ x + y = 12 cm …(i)
Similarly,
y + z = 8 cm …(ii)
and x + z = 10 cm …(iii)
By adding (i), (ii) & (iii)
2(x + y + z) = 30
x + y + z = 15 …[∵ x + y = 12
z = 15 – 12 = 3
Putting the value of z in (ii) & (iii),
y + 3 = 8
y = 8 – 3 = 5
x + 3 = 10
x = 10 – 3 = 7
∴ AD = 7 cm, BE = 5 cm, CF = 3 cm
Q: 15.
The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.
Solution:
The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively.
AB = AC
To prove: BD = CD
Proof: AF = AE ..(i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC …[Given
∴ CD = BD
AB = AC
To prove: BD = CD
Proof: AF = AE ..(i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC …[Given
∴ CD = BD
