Equations of circles which touch both the axes and whose centres are at a distance of `2sqrt(2)` units from origin are

Options

A. `x^(2)+y^(2)+-4x+-4y+4=0`
B. `x^(2)+y^(2)+-2x+-2y+4=0`
C. `x^(2)+y^(2)+-x+-y+4=0`
D. None

Answer

To find the equations of circles that touch both axes and whose centers are at a distance of \(2\sqrt{2}\) units from the origin, we can follow these steps:

### Step 1: Understand the properties of the circle
A circle that touches both axes will have its center at the point \((a, a)\), where \(a\) is the radius of the circle. This is because the distance from the center to the x-axis and y-axis must equal the radius.

### Step 2: Set up the distance condition
The distance from the center \((a, a)\) to the origin \((0, 0)\) is given by the formula:
\[ \text{Distance} = \sqrt{(a - 0)^2 + (a - 0)^2} = \sqrt{2a^2} = a\sqrt{2} \]
According to the problem, this distance is equal to \(2\sqrt{2}\):
\[ a\sqrt{2} = 2\sqrt{2} \]

### Step 3: Solve for \(a\)
To find \(a\), we can divide both sides by \(\sqrt{2}\):
\[ a = 2 \]

### Step 4: Determine the center of the circle
Since \(a = 2\), the center of the circle can be either \((2, 2)\) or \((-2, -2)\) because both points are at a distance of \(2\sqrt{2}\) from the origin.

### Step 5: Write the equations of the circles
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For the center \((2, 2)\) and radius \(2\):
\[ (x - 2)^2 + (y - 2)^2 = 2^2 \implies (x - 2)^2 + (y - 2)^2 = 4 \]
For the center \((-2, -2)\) and radius \(2\):
\[ (x + 2)^2 + (y + 2)^2 = 2^2 \implies (x + 2)^2 + (y + 2)^2 = 4 \]

### Final Equations
Thus, the equations of the circles are:
1. \((x - 2)^2 + (y - 2)^2 = 4\)
2. \((x + 2)^2 + (y + 2)^2 = 4\)