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

Options

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

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:
\[ x^2 + y^2 + Dx + Ey + F = 0 \]
where \(D\), \(E\), and \(F\) are constants.

### Step 2: Substitute the points into the equation
Since the circle passes through the point (0,0), we can substitute \(x = 0\) and \(y = 0\) into the equation:
\[ 0^2 + 0^2 + D(0) + E(0) + F = 0 \implies F = 0 \]
Thus, the equation simplifies to:
\[ x^2 + y^2 + Dx + Ey = 0 \]

### Step 3: Substitute the second point (a,0)
Next, we substitute the point (a,0) into the equation:
\[ a^2 + 0^2 + Da + E(0) = 0 \implies a^2 + Da = 0 \]
This can be rearranged to:
\[ D = -\frac{a^2}{a} = -a \quad (a \neq 0) \]

### Step 4: Substitute the third point (0,b)
Now, we substitute the point (0,b) into the equation:
\[ 0^2 + b^2 + D(0) + Eb = 0 \implies b^2 + Eb = 0 \]
This can be rearranged to:
\[ E = -\frac{b^2}{b} = -b \quad (b \neq 0) \]

### Step 5: Write the equation of the circle with D and E
Now we can substitute \(D\) and \(E\) back into the circle's equation:
\[ x^2 + y^2 - ax - by = 0 \]

### Step 6: Find the center of the circle
The center of the circle can be found using the formula for the center \((-D/2, -E/2)\):
\[ \text{Center} = \left(-\frac{-a}{2}, -\frac{-b}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right) \]

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

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