ABCD is a rectangle wih sides AB=p, BC=q,. If AB and AD are taken negative directions of coordinate axes. then the equaton of the circumscribing the rectangle is

Options

A. a) `x^(2)+y^(2)+px+qy=0`
B. b) `x^(2)+y^(2)-px-qy=0`
C. c) `x^(2)+y^(2)+2px+2qy=0`
D. d) `x^(2)+y^(2)-2px-2qy=0`

Answer

To find the equation of the circle circumscribing the rectangle ABCD with sides AB = p and BC = q, where AB and AD are taken in the negative directions of the coordinate axes, we can follow these steps:

### Step 1: Define the coordinates of the rectangle
Given that AB = p and BC = q, we can place the rectangle in the coordinate system as follows:
- Point A (0, 0) (the origin)
- Point B (-p, 0) (moving p units left on the x-axis)
- Point C (-p, -q) (moving q units down on the y-axis)
- Point D (0, -q) (moving back to the y-axis)

### Step 2: Identify the center of the circumscribing circle
The center of the circumscribing circle of a rectangle is at the midpoint of the diagonal connecting opposite corners. We can find the midpoint of AC (or BD) as follows:
- Midpoint M = \(\left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}\right)\)
- Substituting the coordinates of A and C:
\[ M = \left(\frac{0 + (-p)}{2}, \frac{0 + (-q)}{2}\right) = \left(-\frac{p}{2}, -\frac{q}{2}\right) \]

### Step 3: Calculate the radius of the circumscribing circle
The radius R of the circle can be found using the distance formula from the center to any vertex of the rectangle (let's use point A):
\[ R = \sqrt{\left(-\frac{p}{2} - 0\right)^2 + \left(-\frac{q}{2} - 0\right)^2} = \sqrt{\left(-\frac{p}{2}\right)^2 + \left(-\frac{q}{2}\right)^2} \]
\[ = \sqrt{\frac{p^2}{4} + \frac{q^2}{4}} = \frac{1}{2}\sqrt{p^2 + q^2} \]

### 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 the center M and radius R:
\[ \left(x + \frac{p}{2}\right)^2 + \left(y + \frac{q}{2}\right)^2 = \left(\frac{1}{2}\sqrt{p^2 + q^2}\right)^2 \]
\[ \left(x + \frac{p}{2}\right)^2 + \left(y + \frac{q}{2}\right)^2 = \frac{p^2 + q^2}{4} \]

### Step 5: Expand the equation
Expanding the left side:
\[ \left(x^2 + px + \frac{p^2}{4}\right) + \left(y^2 + qy + \frac{q^2}{4}\right) = \frac{p^2 + q^2}{4} \]
Combining terms:
\[ x^2 + y^2 + px + qy + \frac{p^2 + q^2}{4} - \frac{p^2 + q^2}{4} = 0 \]
Thus, we have:
\[ x^2 + y^2 + px + qy = 0 \]

### Final Answer
The equation of the circle circumscribing the rectangle ABCD is:
\[ x^2 + y^2 + px + qy = 0 \]