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

• A

  $m(m+n)$
• Correct answer
  B

  $m+n$
• C

  $n(m+n)$
• D

  $\frac{1}{2}(m+n)$

Solution

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 $\Rightarrow \mathrm{AB}=\mathrm{MN}=\mathrm{m}+\mathrm{n}$

Hence, the correct answer is option: 'B'.