If a circle passes through the points `(0,0),(a,0)` and `(0,b),` then find the coordinates of its centre.

Options

A. `(a/2,b/2)`
B. `(-a/2,-b/2)`
C. `(a,b)`
D. `"none of these"`

Explanation

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 1: Write the general equation of the circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((g, f)\) are the coordinates of the center of the circle, and \(c\) is a constant.

### Step 2: Substitute the first point (0,0)
Since the circle passes through the point (0,0), we substitute \(x = 0\) and \(y = 0\) into the equation:
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0 \]
Thus, the equation simplifies to:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 3: Substitute the second point (a,0)
Now, we substitute the second point (a,0) into the equation:
\[ a^2 + 0^2 + 2g(a) + 2f(0) = 0 \implies a^2 + 2ga = 0 \]
Factoring out \(a\) (assuming \(a \neq 0\)):
\[ a(a + 2g) = 0 \implies a + 2g = 0 \implies g = -\frac{a}{2} \]

### Step 4: Substitute the third point (0,b)
Next, we substitute the third point (0,b) into the equation:
\[ 0^2 + b^2 + 2g(0) + 2f(b) = 0 \implies b^2 + 2fb = 0 \]
Factoring out \(b\) (assuming \(b \neq 0\)):
\[ b(b + 2f) = 0 \implies b + 2f = 0 \implies f = -\frac{b}{2} \]

### Step 5: Find the coordinates of the center
Now that we have \(g\) and \(f\), we can find the coordinates of the center of the circle:
\[ \text{Center} = (-g, -f) = \left(-\left(-\frac{a}{2}\right), -\left(-\frac{b}{2}\right)\right) = \left(\frac{a}{2}, \frac{b}{2}\right) \]

### Final Answer
The coordinates of the center of the circle are:
\[ \left(\frac{a}{2}, \frac{b}{2}\right) \]