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TS EAMCET 2017 Code A Solved Paper
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© examsnet.com
Question : 11
Total: 160
The radical center of the circles
x
2
+
y
2
−
4
x
−
6
y
+
5
=
0
,
x
2
+
y
2
−
2
x
−
4
y
−
1
=
0
,
x
2
+
y
2
−
6
x
−
2
y
=
0
lies on the line
x
+
y
−
5
=
0
2
x
−
4
y
+
7
=
0
4
x
−
6
y
+
5
=
0
18
x
−
12
y
+
1
=
0
Validate
Solution:
The given equations are
S
1
≡
x
2
+
y
2
−
4
x
−
6
y
+
5
=
0
S
2
≡
x
2
+
y
2
−
2
x
−
4
y
−
1
=
0
S
3
≡
x
2
+
y
2
−
6
x
−
2
y
=
0
Calculate the value of
(
S
1
−
S
2
)
S
1
−
S
2
=
0
2
x
+
2
y
−
6
=
0
x
+
y
−
3
=
0
.......(1)
Calculate the value of
(
S
2
−
S
3
)
S
2
−
S
3
=
0
4
x
−
2
y
−
1
=
0
.....(2)
On solving equation (1) and
(
2
)
,
we get
x
=
7
6
y
=
11
6
The point
(
7
6
,
11
6
)
satisfies the equation
18
x
−
12
y
+
1
=
0
© examsnet.com
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