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VITEEE 2019 Solved Paper
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© examsnet.com
Question : 90
Total: 125
A circle has radius 3 and its centre lies on the line y = x - 1 . The equation of the circle, if it passes through (7,3), is
x
2
+
y
2
+
8
x
−
6
y
+
16
=
0
x
2
+
y
2
−
8
x
+
6
y
+
16
=
0
x
2
+
y
2
−
8
x
−
6
y
−
16
=
0
x
2
+
y
2
−
8
x
−
6
y
+
16
=
0
Validate
Solution:
Let
(
h
,
k
)
be the centre of the circle. Then
k
=
h
−
1
.
Therefore, the equation of the circle is given by
(
x
−
h
)
2
+
[
y
−
(
h
−
1
)
]
2
=
9
.
.
.
. (1)
Given that the circle passes through the point
(
7
,
3
)
and hence we get
(
7
−
h
)
2
+
(
3
−
(
h
−
1
)
)
2
=
9
(
7
−
h
)
2
+
(
4
−
h
)
2
=
9
h
2
−
11
h
+
28
=
0
which gives
(
h
−
7
)
(
h
−
4
)
=
0
⇒
h
=
4
or
h
=
7
Therefore, the required equations of the circles are
x
2
+
y
2
−
8
x
−
6
y
+
16
=
0
or,
x
2
+
y
2
−
14
x
−
12
y
+
76
=
0
© examsnet.com
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