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JEE Main 2012
19 May
Online
Exam
JEE Main
Year
2012
Date
19 May
Mode
Online
Subject
Mathematics
Chapter
Circle
JEE Main 2012 · 19 May · Online · Mathematics · Circle
If the line y=m x+1 meets the circle x^2+y^2+3 x=0 in two points...
If the line
y
=
m
x
+
1
y=m x+1
y
=
m
x
+
1
meets the circle
x
2
+
y
2
+
3
x
=
0
x^2+y^2+3 x=0
x
2
+
y
2
+
3
x
=
0
in two points equidistant from and on opposite sides of
x
x
x
-axis, then
Options
A
3
m
+
2
=
0
3 m+2=0
3
m
+
2
=
0
B
3
m
−
2
=
0
3 m-2=0
3
m
−
2
=
0
C
2
m
+
3
=
0
2 m+3=0
2
m
+
3
=
0
D
2
m
−
3
=
0
2 m-3=0
2
m
−
3
=
0
Correct answer
B.
3
m
−
2
=
0
3 m-2=0
3
m
−
2
=
0
Step-by-step solution
Circle:
x
2
+
y
2
+
3
x
=
0
x^2+y^2+3 x=0
x
2
+
y
2
+
3
x
=
0
Centre,
B
=
B=
B
=
\left(-\frac{3}{2},0\right)
<
i
m
g
s
r
c
=
"
/
a
p
i
/
p
r
o
x
y
−
i
m
a
g
e
?
u
r
l
=
h
t
t
p
s
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<
im
g
src
=
"/
a
p
i
/
p
ro
x
y
−
ima
g
e
?
u
r
l
=
h
ttp
s
y
−
i
n
t
e
r
c
e
p
t
o
f
t
h
e
l
i
n
e
-intercept of the line
−
in
t
erce
pt
o
f
t
h
e
l
in
e
=1
∴
\therefore \quad
∴
\mathrm{A}=(0,1) Slope of line,
m
=
tan
θ
=
O
A
O
B
m=\tan \theta=\frac{O A}{O B}
m
=
tan
θ
=
OB
O
A
⇒
m
=
1
3
2
=
2
3
⇒
3
m
−
2
=
0
\begin{aligned} & \Rightarrow m=\frac{1}{\frac{3}{2}}=\frac{2}{3}\\ & \Rightarrow 3 m-2=0 \end{aligned}
⇒
m
=
2
3
1
=
3
2
⇒
3
m
−
2
=
0
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