Equation of the circle which passes through the origin and cuts off intercepts of length 2 units from negative co-ordinate axes , is

Options

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

Answer

To find the equation of the circle that passes through the origin and cuts off intercepts of length 2 units from the negative coordinate axes, we can follow these steps:

### Step 1: Understand the Circle's Properties
The circle passes through the origin (0, 0) and intercepts the negative x-axis and negative y-axis at points that are 2 units away from the origin. This means the intercepts on the negative axes are (-2, 0) and (0, -2).

### Step 2: General Equation of the Circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Since the circle passes through the origin (0, 0), we can substitute these coordinates into the equation to find \(c\):
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0 \]
Thus, the equation simplifies to:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 3: Determine the Intercepts
The circle cuts off intercepts of length 2 units from the negative axes. The x-intercept and y-intercept can be found using the following formulas:
- For the x-axis: \(2 \sqrt{g^2 - c} = 2\)
- For the y-axis: \(2 \sqrt{f^2 - c} = 2\)

Since \(c = 0\), we have:
\[ 2 \sqrt{g^2} = 2 \implies \sqrt{g^2} = 1 \implies |g| = 1 \]
\[ 2 \sqrt{f^2} = 2 \implies \sqrt{f^2} = 1 \implies |f| = 1 \]

### Step 4: Analyze the Signs of \(g\) and \(f\)
Since the intercepts are on the negative axes, the center of the circle, which is at \((-g, -f)\), must lie in the third quadrant. This means both \(-g < 0\) and \(-f < 0\), implying:
\[ g > 0 \quad \text{and} \quad f > 0 \]
Thus, \(g\) and \(f\) must both be equal to \(1\) (since they cannot be negative).

### Step 5: Substitute Values of \(g\) and \(f\)
Now substituting \(g = 1\) and \(f = 1\) into the equation:
\[ x^2 + y^2 + 2(1)x + 2(1)y = 0 \]
This simplifies to:
\[ x^2 + y^2 + 2x + 2y = 0 \]

### Step 6: Final Equation
The final equation of the circle is:
\[ x^2 + y^2 + 2x + 2y = 0 \]

### Conclusion
Thus, the equation of the circle that passes through the origin and cuts off intercepts of length 2 units from the negative coordinate axes is:
\[ x^2 + y^2 + 2x + 2y = 0 \]