If `(alpha, beta)` is the centre of a circle passing through the origin, then its equation is

Options

A. `x ^(2) + y^(2) -alpha x - beta y =0`
B. ` x ^(2) + y^(2) + 2 alpha x + 2 betay =0`
C. `x ^(2) +y^(2) -2 alpha x - 2 beta y =0`
D. `x ^(2) + y^(2) + alpha x + beta y =0`

Answer

To find the equation of a circle with center \((\alpha, \beta)\) that passes through the origin \((0, 0)\), we can follow these steps:

### Step 1: Understand the standard equation of a circle
The standard equation of a circle with center \((h, k)\) and radius \(R\) is given by:
\[ (x - h)^2 + (y - k)^2 = R^2 \]
In our case, the center is \((\alpha, \beta)\), so we can rewrite the equation as:
\[ (x - \alpha)^2 + (y - \beta)^2 = R^2 \]

### Step 2: Determine the radius \(R\)
Since the circle passes through the origin \((0, 0)\), we can find the radius \(R\) by calculating the distance from the center \((\alpha, \beta)\) to the origin. The distance formula is:
\[ R = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (\alpha, \beta)\), we get:
\[ R = \sqrt{(\alpha - 0)^2 + (\beta - 0)^2} = \sqrt{\alpha^2 + \beta^2} \]

### Step 3: Substitute \(R\) back into the circle equation
Now we substitute \(R^2\) into the equation of the circle:
\[ (x - \alpha)^2 + (y - \beta)^2 = (\sqrt{\alpha^2 + \beta^2})^2 \]
This simplifies to:
\[ (x - \alpha)^2 + (y - \beta)^2 = \alpha^2 + \beta^2 \]

### Step 4: Expand the equation
Now we expand the left-hand side:
\[ (x - \alpha)^2 + (y - \beta)^2 = x^2 - 2\alpha x + \alpha^2 + y^2 - 2\beta y + \beta^2 \]
Thus, we have:
\[ x^2 - 2\alpha x + \alpha^2 + y^2 - 2\beta y + \beta^2 = \alpha^2 + \beta^2 \]

### Step 5: Simplify the equation
Subtract \(\alpha^2 + \beta^2\) from both sides:
\[ x^2 - 2\alpha x + \alpha^2 + y^2 - 2\beta y + \beta^2 - \alpha^2 - \beta^2 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2\alpha x - 2\beta y = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 2\alpha x - 2\beta y = 0 \]