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Byju's Answer
Standard XII
Mathematics
Direct Common Tangent
Tangents are ...
Question
T
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n
t
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f
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m
t
h
e
p
o
i
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t
(
h
,
k
)
t
o
t
h
e
c
i
r
c
l
e
x
2
+
y
2
=
a
2
;
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t
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r
p
o
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s
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f
c
o
n
t
a
c
t
i
s
a
(
h
2
+
k
2
−
a
2
)
3
2
h
2
+
k
2
.
A
True
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B
False
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Solution
The correct option is
A
True
Let tangents are drawn from P(h,k) to circle
x
2
+
y
2
=
a
2
at point of contact of tangents be Q and R
Equation of chord of contact QR is
h
x
+
k
y
=
a
2
h
x
+
k
y
−
a
2
=
0
−
(
1
)
Draw PS perpendicular to QR
PS=
h
2
+
k
2
−
a
2
√
h
2
+
k
2
length of tangent =PR=
√
S
1
⇒PR=
√
h
2
+
k
2
−
a
2
In △PSR ,
P
S
2
+
S
R
2
=
P
R
2
⇒
S
R
2
=
(
√
h
2
+
k
2
−
a
2
)
2
−
(
h
2
+
k
2
−
a
2
√
h
2
+
k
2
)
2
⟹
S
R
2
=
(
h
2
+
k
2
−
a
2
)
(
1
−
h
2
+
k
2
−
a
2
h
2
+
k
2
⟹
S
R
2
=
(
h
2
+
k
2
−
a
2
)
(
h
2
+
k
2
−
h
2
−
k
2
+
a
2
h
2
+
k
2
⟹
S
R
2
=
(
h
2
+
k
2
−
a
2
)
(
a
2
h
2
+
k
2
⟹
S
R
2
=
a
√
(
h
2
+
k
2
−
a
2
)
√
h
2
+
k
2
But QR=2SR
⟹
Q
R
=
2
a
√
h
2
+
k
2
−
a
2
√
h
2
+
k
2
Area of △PQR=
1
2
(
Q
R
)
(
P
S
)
=
1
2
×
2
a
√
h
2
+
k
2
−
a
2
√
h
2
+
k
2
×
h
2
+
k
2
−
a
2
√
h
2
+
k
2
Area=
a
(
h
2
+
k
2
−
a
2
)
3
2
h
2
+
k
2
Suggest Corrections
0
Similar questions
Q.
Tangents are drawn from the point
(
h
,
k
)
to the circle
x
2
+
y
2
=
a
2
; prove that the area of the triangle formed by them and the straight line joining their points of contact is
a
(
h
2
+
k
2
−
a
2
)
3
2
h
2
+
k
2
.
Q.
The curves
x
2
a
2
+
k
1
+
y
2
b
2
+
k
1
=
1
,
x
2
a
2
+
k
2
+
y
2
b
2
+
k
2
=
1
where
k
1
≠
k
2
intersect at an angle
Q.
If
h
2
+
k
2
>
a
2
, then the number of solutions of the equation
h
x
+
k
y
=
a
2
and
x
2
+
y
2
=
a
2
is
Q.
The angle between the curves
x
2
a
2
+
k
1
+
y
2
b
2
+
k
1
=
1
and
x
2
a
2
+
k
2
+
y
2
b
2
+
k
2
=
1
is
Q.
Let the equation of a circle be
x
2
+
y
2
=
a
2
lf
h
2
+
k
2
−
a
2
<
0
, then the line
h
x
+
k
y
=
a
2
is the
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