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A straight line meets the co-ordinate axes at $\mathrm{A}$ and $\mathrm{B}$. A circle is circumscribed about the triangle $\mathrm{OAB}, \mathrm{O}$ being the origin. If $\mathrm{m}$ and $\mathrm{n}$ are the distances of the tangent to the circle at the origin from the points $\mathrm{A}$ and $\mathrm{B}$ respectively, the diameter of the circle is
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The correct answer is:
$m+n$
Clearly, AB is one of diameter
$\because \mathrm{AM}$ and $\mathrm{BN}$ are parallel and $\angle \mathrm{BNO}=\angle \mathrm{AMO}=\pi / 2$
$\therefore$ Points $\mathrm{N}, \mathrm{O}$ and $\mathrm{M}$ are collinear.
$\therefore \triangle \mathrm{BNMA}$ is a rectangle
$\Rightarrow \mathrm{AB}=\mathrm{MN}=\mathrm{m}+\mathrm{n}$
$\because \mathrm{AM}$ and $\mathrm{BN}$ are parallel and $\angle \mathrm{BNO}=\angle \mathrm{AMO}=\pi / 2$
$\therefore$ Points $\mathrm{N}, \mathrm{O}$ and $\mathrm{M}$ are collinear.
$\therefore \triangle \mathrm{BNMA}$ is a rectangle
$\Rightarrow \mathrm{AB}=\mathrm{MN}=\mathrm{m}+\mathrm{n}$
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