If a circle passes through the point (0,0), (a,0) and (0,b), then the coordinates of its centre are

Options

A. (a,b)
B. `(-(a)/(2),-(b)/(2))`
C. `((a)/(2),(b)/(2))`
D. (-a, -b)

Answer

To find the coordinates of the center of a circle that passes through the points (0,0), (a,0), and (0,b), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Equation of a Circle**:
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

2. **Substituting the First Point (0,0)**:
Substitute the point (0,0) into the circle equation:
\[ (0 - h)^2 + (0 - k)^2 = r^2 \]
This simplifies to:
\[ h^2 + k^2 = r^2 \quad \text{(Equation 1)} \]

3. **Substituting the Second Point (a,0)**:
Substitute the point (a,0) into the circle equation:
\[ (a - h)^2 + (0 - k)^2 = r^2 \]
Expanding this gives:
\[ (a^2 - 2ah + h^2 + k^2) = r^2 \]
Since \(h^2 + k^2 = r^2\) from Equation 1, we can replace \(r^2\):
\[ a^2 - 2ah + r^2 = r^2 \]
This simplifies to:
\[ a^2 - 2ah = 0 \quad \text{(Equation 2)} \]

4. **Substituting the Third Point (0,b)**:
Substitute the point (0,b) into the circle equation:
\[ (0 - h)^2 + (b - k)^2 = r^2 \]
Expanding this gives:
\[ (h^2 + (b^2 - 2bk + k^2)) = r^2 \]
Again, using \(h^2 + k^2 = r^2\):
\[ h^2 + b^2 - 2bk + r^2 = r^2 \]
This simplifies to:
\[ b^2 - 2bk = 0 \quad \text{(Equation 3)} \]

5. **Solving Equations 2 and 3**:
From Equation 2:
\[ a^2 - 2ah = 0 \implies a(a - 2h) = 0 \]
This gives us two possibilities: \(a = 0\) or \(h = \frac{a}{2}\).

From Equation 3:
\[ b^2 - 2bk = 0 \implies b(b - 2k) = 0 \]
This gives us two possibilities: \(b = 0\) or \(k = \frac{b}{2}\).

6. **Determining the Center Coordinates**:
Since the circle passes through the origin (0,0), the cases where \(a = 0\) or \(b = 0\) are not valid (as they would imply the circle is degenerate). Thus, we have:
\[ h = \frac{a}{2} \quad \text{and} \quad k = \frac{b}{2} \]

7. **Final Result**:
Therefore, the coordinates of the center of the circle are:
\[ \left(\frac{a}{2}, \frac{b}{2}\right) \]