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Mathematics
Mathematics
Problem
If
a
>
2
b
>
0
a > 2b > 0
a
>
2
b
>
0
, then the positive value of
m
m
m
for which
y
=
m
x
−
b
1
+
m
2
y = mx - b\sqrt{1+m^2}
y
=
m
x
−
b
1
+
m
2
is a common tangent to
x
2
+
y
2
=
b
2
x^2 + y^2 = b^2
x
2
+
y
2
=
b
2
and
(
x
−
a
)
2
+
y
2
=
b
2
(x-a)^2 + y^2 = b^2
(
x
−
a
)
2
+
y
2
=
b
2
is a.
2
b
(
a
2
−
4
b
2
)
\frac{2b}{\sqrt{(a^2 - 4b^2)}}
(
a
2
−
4
b
2
)
2
b
b.
(
a
2
−
4
b
2
)
2
b
\frac{\sqrt{(a^2 - 4b^2)}}{2b}
2
b
(
a
2
−
4
b
2
)
c.
2
b
a
−
2
b
\frac{2b}{a - 2b}
a
−
2
b
2
b
d.
b
a
−
2
b
\frac{b}{a - 2b}
a
−
2
b
b
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