Let $\mathrm{A}\mathrm{}\left(\mathrm{2,0}\right)$ and $\mathrm{B}\left(-\mathrm{2,0}\right)$ are two fixed vertices of $∆\mathrm{A}\mathrm{B}\mathrm{C}.$ If the vertex $\mathrm{C}\mathrm{}$ moves in the first quadrant in such a way that $\cot \mathrm{A}+\cot \mathrm{B}=\mathrm{2,}\mathrm{}$ then the locus of the point $\mathrm{C}$ is 

• Correct answer
  A

  $\mathrm{y}=2\mathrm{}$
• B

  $\mathrm{x}=4\mathrm{}$
• C

  $\mathrm{x}=2\mathrm{}$
• D

  $\mathrm{y}=1\mathrm{}$

Solution

Given that, coordinates of two fixed points $\mathrm{A}\mathrm{}$ and $\mathrm{B}\mathrm{}$ are $\left(2,\mathrm{}0\right)$ and $\left(-2,\mathrm{}0\right)$ respectively
Let variable point $\mathrm{C}\mathrm{}$ is $\left(\mathrm{h},\mathrm{k}\right).$ 

From the adjoining figure,
$\cot \mathrm{A}=\frac{\mathrm{D}\mathrm{A}}{\mathrm{C}\mathrm{D}}=\frac{2-\mathrm{h}}{\mathrm{k}}$
$\cot \mathrm{B}=\frac{\mathrm{B}\mathrm{D}}{\mathrm{C}\mathrm{D}}=\frac{2+\mathrm{h}}{\mathrm{k}}$
But $\cot \mathrm{A}+\cot \mathrm{B}=2,$ so we have
$\mathrm{}\frac{2-\mathrm{h}}{\mathrm{k}}+\frac{2+\mathrm{h}}{\mathrm{k}}=2\Rightarrow \frac{2}{\mathrm{k}}=1$
Hence, locus of $\mathrm{C}$ is $\mathrm{y}=2$

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