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JEE Main Math Class 11 Coordinate Geometry Part 3 Questions
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© examsnet.com
Question : 8
Total: 100
Two circles each of radius 5 units touch each other at the point
(
1
,
2
)
. If the equation of their common tangent is
4
x
+
3
y
=
10
and
C
1
(
α
,
β
)
and
C
2
(
γ
,
δ
)
,
C
1
≠
C
2
are their centres, then
|
(
α
+
β
)
(
γ
+
δ
)
|
is equal to
[2021, 27 Aug. Shift-II]
Your Answer:
Validate
Solution:
Equation of circles
(
x
−
1
)
2
+
(
y
−
2
)
2
+
λ
(
4
x
+
3
y
−
10
)
=
0
x
2
+
y
2
+
(
2
λ
−
1
)
2
x
+
(
3
2
λ
−
2
)
2
y
+
5
−
10
λ
=
0
r
=
√
(
2
λ
−
1
)
2
+
(
3
2
λ
−
2
)
2
−
5
+
10
λ
=
5
⇒
4
λ
2
+
1
−
4
λ
+
9
4
λ
2
+
4
−
6
λ
−
5
+
10
λ
=
25
⇒
25
4
λ
2
−
25
=
0
⇒
λ
=
±
2
For
λ
=
2
x
2
+
y
2
+
6
x
+
2
y
−
15
=
0
,
C
1
(
−
3
,
−
1
)
For
λ
=
−
2
x
2
+
y
2
−
10
x
−
10
y
+
25
=
0
,
C
2
(
5
,
5
)
|
(
α
+
β
)
(
γ
+
δ
)
|
=
|
−
4
×
10
|
=
40
© examsnet.com
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