To find the coordinates of the center of the circle passing through the points $$(0,0), (x, 0)$$, and $$(0, y)$$, we can follow these steps.

First, let's understand that the general equation of a circle with center $$(h, k)$$ and radius $$r$$ can be written as:

$$ (x - h)^2 + (y - k)^2 = r^2 $$

Since the circle passes through the origin $$(0,0)$$, substituting $$(0,0)$$ in the above equation, we get:

$$ h^2 + k^2 = r^2 \quad \text{(1)} $$

Next, since the circle passes through $$(x,0)$$, substituting $$(x,0)$$ in the circle equation, we get:

$$ (x - h)^2 + k^2 = r^2 \quad \text{(2)} $$

Substituting $$r^2$$ from equation (1) into equation (2) gives:

$$ (x - h)^2 + k^2 = h^2 + k^2 $$

Canceling $$k^2$$ from both sides, we get:

$$ (x - h)^2 = h^2 $$

Solving for $$h$$, we get two possible solutions:

$$ x - h = h \quad \text{or} \quad x - h = -h $$

$$ x = 2h \quad \text{or} \quad x = 0 $$

Since $$x \neq 0$$ in our case, we have:

$$ h = \frac{x}{2} $$

Similarly, since the circle passes through $$(0, y)$$, substituting $$(0, y)$$ into the circle equation, we get:

$$ h^2 + (y - k)^2 = r^2 \quad \text{(3)} $$

Substituting $$r^2$$ from equation (1) into equation (3) gives:

$$ h^2 + (y - k)^2 = h^2 + k^2 $$

Canceling $$h^2$$ from both sides, we get:

$$ (y - k)^2 = k^2 $$

Solving for $$k$$, we get two possible solutions:

$$ y - k = k \quad \text{or} \quad y - k = -k $$

$$ y = 2k \quad \text{or} \quad y = 0 $$

Since $$y \neq 0$$ in our case, we have:

$$ k = \frac{y}{2} $$

Therefore, the coordinates of the center of the circle are:

$$ \left(\frac{x}{2}, \frac{y}{2}\right) $$

So, the correct answer is:

Option B

$$ \left(\frac{x}{2}, \frac{y}{2}\right) $$

Circle

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