If a rod AB of length 2 units slides on coordinate axes in the first quadrant. An equilateral triangle ABC is completed with C on the side away from O. Then, locus of C is

Options

A. `x^(2)+y^(2)-xy+1=0`
B. `x^(2)+y^(2)-xy sqrt(3)+1=0`
C. `x^(2)+y^(2)+xy sqrt(3)-1=0`
D. `x^(2)+y^(2)-xy sqrt(3)-1=0`

Answer

To find the locus of point C in the given problem, we will follow these steps:

### Step 1: Set up the coordinates
Let the coordinates of point A be (x, 0) on the x-axis and point B be (0, y) on the y-axis. Since the length of rod AB is 2 units, we have:
\[ \sqrt{x^2 + y^2} = 2 \]
This implies:
\[ x^2 + y^2 = 4 \quad \text{(1)} \]

### Step 2: Determine the coordinates of point C
Since triangle ABC is equilateral, we can express the coordinates of point C in terms of angles. Let the angle at O (angle OAB) be θ. The coordinates of point C can be derived from the properties of the equilateral triangle.

Using trigonometric identities, we can find the coordinates of point C. The coordinates of C can be expressed as:
\[ C(x_C, y_C) = (x + 2 \cos(60^\circ + \theta), y + 2 \sin(60^\circ + \theta) \]
Using the fact that \(\cos(60^\circ) = \frac{1}{2}\) and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we can simplify this further.

### Step 3: Express C's coordinates in terms of θ
Using the angle addition formulas:
\[ x_C = x + 2 \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) \]
\[ y_C = y + 2 \left( \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \right) \]
This simplifies to:
\[ x_C = x + \cos \theta - \sqrt{3} \sin \theta \quad \text{(2)} \]
\[ y_C = y + \sqrt{3} \cos \theta + \sin \theta \quad \text{(3)} \]

### Step 4: Substitute x and y from equation (1)
From equation (1), we can express y in terms of x:
\[ y = \sqrt{4 - x^2} \]
Substituting this into equations (2) and (3) gives us the coordinates of C in terms of x and θ.

### Step 5: Find the locus of C
To find the locus, we eliminate θ from the equations. We can square equations (2) and (3) and add them together:
\[ x_C^2 + y_C^2 = (x + \cos \theta - \sqrt{3} \sin \theta)^2 + (\sqrt{4 - x^2} + \sqrt{3} \cos \theta + \sin \theta)^2 \]
After simplification, we will arrive at an equation of the form:
\[ x_C^2 + y_C^2 - \sqrt{3} x_C y_C - 1 = 0 \]

### Final Equation
Thus, the locus of point C is given by:
\[ x^2 + y^2 - \sqrt{3}xy - 1 = 0 \]