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Mathematics
The equations of the tangents to the circle x2 + y2 - 6x + 4y = 12 , which are parallel to the straight line 4x + 3y + 5 = 0 , are
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Q. The equations of the tangents to the circle
x
2
+
y
2
−
6
x
+
4
y
=
12
, which are parallel to the straight line
4
x
+
3
y
+
5
=
0
, are
MHT CET
MHT CET 2010
A
3
x
−
4
y
−
1
9
=
0
,
3
x
−
4
y
+
31
=
0
B
4
x
+
3
y
−
19
=
0
,
4
x
+
3
y
+
31
=
0
C
4
x
+
3
y
+
19
=
0
,
4
x
+
3
y
−
31
=
0
D
3
x
−
4
y
+
19
=
0
,
3
x
−
4
y
+
31
=
0
Solution:
Let equation of tangent be
4
x
+
3
y
+
k
=
0
centre of the circle is
(
3
,
−
2
)
.
Then,
√
9
+
4
+
12
=
|
4
(
3
)
+
3
(
−
2
)
+
k
√
16
+
9
|
⇒
6
+
k
=
±
25
⇒
k
=
19
and
−
31
Hence, the tangent are
4
x
+
3
y
+
19
=
0
and
4
x
+
3
y
−
31
=
0