From a circle of radius a, an isosceles right angles triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of mass of the remaining portion from the centre of the circle is

Options

A. `3(pi-1)a`
B. `((pi-1)a)/(6)`
C. `(a)/(3(pi-1))`
D. `(a)/(3(pi+1))`

Explanation

To solve the problem of finding the distance of the center of mass of the remaining portion from the center of the circle after removing an isosceles right triangle with the hypotenuse as the diameter of the circle, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Geometry**:
- We have a circle with radius \( a \). The diameter of the circle serves as the hypotenuse of the isosceles right triangle.
- The triangle will have its base along the diameter of the circle and its apex at the topmost point of the circle.

2. **Determine the Area and Mass of the Circle**:
- The area \( A_c \) of the circle is given by:
\[ A_c = \pi a^2 \]
- Assuming uniform mass density \( \sigma \), the mass \( M \) of the circle can be expressed as:
\[ M = \sigma A_c = \sigma \pi a^2 \]

3. **Determine the Area and Mass of the Triangle**:
- The base of the triangle is the diameter of the circle, which is \( 2a \), and the height is \( a \).
- The area \( A_t \) of the triangle is:
\[ A_t = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2a \times a = a^2 \]
- The mass \( M_t \) of the triangle is:
\[ M_t = \sigma A_t = \sigma a^2 \]

4. **Determine the Center of Mass of the Triangle**:
- The center of mass of the isosceles right triangle lies at a distance of \( \frac{1}{3} \) of the height from the base. The height of the triangle is \( a \), so the center of mass is located at:
\[ y_t = \frac{a}{3} \]
- The coordinates of the center of mass of the triangle (taking the base along the x-axis) are \( (0, \frac{a}{3}) \).

5. **Set Up the Center of Mass Equation**:
- The center of mass of the remaining portion (circle minus triangle) can be found using the formula:
\[ y_{cm} = \frac{M y_c - M_t y_t}{M - M_t} \]
- Here, \( y_c = 0 \) (the center of the circle), and \( y_t = \frac{a}{3} \):
\[ y_{cm} = \frac{M \cdot 0 - M_t \cdot \frac{a}{3}}{M - M_t} = \frac{-M_t \cdot \frac{a}{3}}{M - M_t} \]

6. **Substituting Values**:
- Substitute \( M = \sigma \pi a^2 \) and \( M_t = \sigma a^2 \):
\[ y_{cm} = \frac{-\sigma a^2 \cdot \frac{a}{3}}{\sigma \pi a^2 - \sigma a^2} \]
- Simplifying gives:
\[ y_{cm} = \frac{-\frac{\sigma a^3}{3}}{\sigma a^2 (\pi - 1)} = \frac{-a}{3(\pi - 1)} \]

7. **Distance from the Center**:
- The distance of the center of mass of the remaining portion from the center of the circle is:
\[ d = \left| y_{cm} \right| = \frac{a}{3(\pi - 1)} \]

### Final Answer
The distance of the center of mass of the remaining portion from the center of the circle is:
\[ \frac{a}{3(\pi - 1)} \]