ABCD is a square with side a. If AB and AD are taken as positive coordinate axes then equation of circle circumscribing the square is

Options

A. `x^(2)+y^(2)-ax-ay=0`
B. `x^(2)+y^(2)+ax+ay=0`
C. `x^(2)+y^(2)-ax+ay=0`
D. `x^(2)+y^(2)+ax-ay=0`

Answer

To find the equation of the circle circumscribing the square ABCD with side length \( a \), where \( AB \) and \( AD \) are taken as positive coordinate axes, we can follow these steps:

### Step 1: Identify the coordinates of the square's vertices
Since \( AB \) and \( AD \) are on the positive axes, we can place the square in the coordinate plane as follows:
- \( A(0, 0) \)
- \( B(a, 0) \)
- \( C(a, a) \)
- \( D(0, a) \)

### Step 2: Find the center of the circle
The center of the circle that circumscribes the square is the midpoint of the diagonal connecting opposite corners. The coordinates of the center \( O \) can be calculated as:
\[ O\left(\frac{0 + a}{2}, \frac{0 + a}{2}\right) = O\left(\frac{a}{2}, \frac{a}{2}\right) \]

### Step 3: Calculate the radius of the circle
The radius \( R \) of the circumscribing circle is the distance from the center \( O \) to any vertex of the square. We can use the distance formula to find this:
\[ R = \sqrt{\left(a - \frac{a}{2}\right)^2 + \left(0 - \frac{a}{2}\right)^2} \]
Calculating this gives:
\[ R = \sqrt{\left(\frac{a}{2}\right)^2 + \left(-\frac{a}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{2a^2}{4}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}} \]

### Step 4: Write the equation of the circle
The general equation of a circle with center \( (h, k) \) and radius \( R \) is given by:
\[ (x - h)^2 + (y - k)^2 = R^2 \]
Substituting \( h = \frac{a}{2} \), \( k = \frac{a}{2} \), and \( R = \frac{a}{\sqrt{2}} \) into the equation:
\[ \left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 = \left(\frac{a}{\sqrt{2}}\right)^2 \]
This simplifies to:
\[ \left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 = \frac{a^2}{2} \]

### Step 5: Expand the equation
Expanding the left side:
\[ \left(x^2 - ax + \frac{a^2}{4}\right) + \left(y^2 - ay + \frac{a^2}{4}\right) = \frac{a^2}{2} \]
Combining terms gives:
\[ x^2 + y^2 - ax - ay + \frac{a^2}{2} = \frac{a^2}{2} \]
Subtracting \( \frac{a^2}{2} \) from both sides results in:
\[ x^2 + y^2 - ax - ay = 0 \]

### Final Result
Thus, the equation of the circle circumscribing the square ABCD is:
\[ x^2 + y^2 - ax - ay = 0 \]

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