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AIEEE 2004 Math Solved Paper
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© examsnet.com
Question : 24
Total: 68
If the lines
2
x
+
3
y
+
1
+
0
and
3
x
−
y
−
4
=
0
lie along diameter of a circle of circumference
10
π
, then the equation of the circle is
x
2
+
y
2
+
2
x
−
2
y
−
23
=
0
x
2
+
y
2
−
2
x
−
2
y
−
23
=
0
x
2
+
y
2
+
2
x
+
2
y
−
23
=
0
x
2
+
y
2
−
2
x
+
2
y
−
23
=
0
Validate
Solution:
Two diameters are along
2
x
+
3
y
+
1
=
0
and
3
x
−
y
−
4
=
0
solving we get center
(
1
,
−
1
)
circumference
=
2
π
r
=
10
π
∴
r
=
5
.
Required circle is,
(
x
−
1
)
2
+
(
y
+
1
)
2
=
5
2
⇒
x
2
+
y
2
−
2
x
+
2
y
−
23
=
0
© examsnet.com
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