Position vector of a particle moving in `x-y` plane at time `t` is `r=a(1- cos omega t)hat(i)+a sin omega t hat(j)`. The path of the particle is

Options

A. a circle of radius a and centre and (a,0)
B. a circle of radius a and centre at (0,0)
C. an ellipse
D. Neither a circle nor an ellipse

Answer

To solve the problem, we need to analyze the given position vector of the particle moving in the `x-y` plane, which is defined as:

\[ \mathbf{r} = a(1 - \cos(\omega t)) \hat{i} + a \sin(\omega t) \hat{j} \]

### Step 1: Identify the components of the position vector
We can separate the position vector into its `x` and `y` components:

- The `x` component is:
\[ x = a(1 - \cos(\omega t)) \]

- The `y` component is:
\[ y = a \sin(\omega t) \]

### Step 2: Express `cos(ωt)` and `sin(ωt)` in terms of `x` and `y`
From the `x` component, we can express `cos(ωt)`:
\[ \cos(\omega t) = 1 - \frac{x}{a} \]

From the `y` component, we can express `sin(ωt)`:
\[ \sin(\omega t) = \frac{y}{a} \]

### Step 3: Use the Pythagorean identity
We know that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Applying this identity to our expressions, we substitute `sin(ωt)` and `cos(ωt)`:
\[ \left(\frac{y}{a}\right)^2 + \left(1 - \frac{x}{a}\right)^2 = 1 \]

### Step 4: Simplify the equation
Now, we will simplify the equation:
\[ \frac{y^2}{a^2} + \left(1 - \frac{x}{a}\right)^2 = 1 \]
Expanding the second term:
\[ \frac{y^2}{a^2} + \left(1 - \frac{2x}{a} + \frac{x^2}{a^2}\right) = 1 \]
Combining terms:
\[ \frac{y^2}{a^2} + 1 - \frac{2x}{a} + \frac{x^2}{a^2} = 1 \]
Subtracting `1` from both sides:
\[ \frac{y^2}{a^2} - \frac{2x}{a} + \frac{x^2}{a^2} = 0 \]

### Step 5: Multiply through by \(a^2\)
To eliminate the denominators, we multiply the entire equation by \(a^2\):
\[ y^2 - 2ax + x^2 = 0 \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ x^2 - 2ax + y^2 = 0 \]
This can be recognized as the equation of a circle.

### Step 7: Completing the square
To find the center and radius, we complete the square for the `x` terms:
\[ (x - a)^2 + y^2 = a^2 \]

### Conclusion
This represents a circle with:
- Center at \((a, 0)\)
- Radius \(a\)

Thus, the path of the particle is a circle of radius \(a\) centered at \((a, 0)\).

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