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10-1.Circle and System of Circles
hard
If $a > 2b > 0$ then the positive value of m for which $y = mx - b\sqrt {1 + {m^2}} $ is a common tangent to ${x^2} + {y^2} = {b^2}$ and ${(x - a)^2} + {y^2} = {b^2}$, is
A
$\frac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}$
B
$\frac{{\sqrt {{a^2} - 4{b^2}} }}{{2b}}$
C
$\frac{{2b}}{{a - 2b}}$
D
$\frac{b}{{a - 2b}}$
(IIT-2002)
Solution
(a) Any tangent to ${x^2} + {y^2} = {b^2}$ is
$y = mx – b\,\sqrt {1 + {m^2}} .$
It touches ${(x – a)^2} + {y^2} = {b^2}$,
if $\frac{{ma – b\sqrt {1 + {m^2}} }}{{\sqrt {{m^2} + 1} }} = b$ or $ma = 2b\sqrt {1 + {m^2}}$
or ${m^2}{a^2} = 4{b^2} + 4{b^2}{m^2}$,
$\therefore $ $m = \pm \,\frac{{2b}}{{\sqrt {{a^2} – 4{b^2}} }}$.
Standard 11
Mathematics
