Question

The locus of the midpoint of the chord of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}-2=0$ which makes an angle of $120°$ at the center, is

• A. ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}-1=0$
• B. ${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}+\mathrm{y}-1=0$
• C. ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}+1=0$
• D. None of these

Answer 1

The correct option is C

${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}+1=0$

Explanation of the correct option.

Compute the locus:

Given : ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}-2=0$

It is a circle with a radius of $2$ and center at $\mathrm{O}(1,1)$.

Let the midpoint of the chord is $\mathrm{M}(\mathrm{h},\mathrm{k})$.

Let's plot the figure.

From the figure,

$$\begin{gathered} \mathrm{AM}=\mathrm{OA}\sin \left(\mathrm{\theta }\right) \\ \Rightarrow \mathrm{AM}=2\frac{\sqrt{3}}{2} \\ \Rightarrow \mathrm{AM}=\sqrt{3} \end{gathered}$$

Since ${\mathrm{OM}}^2={\mathrm{OA}}^2-{\mathrm{AM}}^2$

$\Rightarrow$ ${\mathrm{OM}}^2=4-{\left(\sqrt{3}\right)}^2$

$\Rightarrow$ ${\mathrm{OM}}^2=1$

From the distance formula, ${\mathrm{OM}}^2={\left(\mathrm{h}-1\right)}^2+{\left(\mathrm{k}-1\right)}^2$ $...\left[distance=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\right]$

Therefore, the locus of $\mathrm{M}(\mathrm{h},\mathrm{k})$ is

$$\begin{gathered} {\left(\mathrm{x}-1\right)}^2+{(y-1)}^2=1 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}+1=0 \end{gathered}$$

Hence, option $\mathrm{C}$ is the correct answer.