The equation of the circle which touches both axes and whose centre is `(x_(1), y _(1))` is

Options

A. `x ^(2) + y^(2) + 2x _(1)(x+y) + x_(1)^(2) =0`
B. `x ^(2) +y^(2) -2x _(1) (x+y) + x_1^(2) =0`
C. ` x ^(2) + y^(2) =x_(1)^(2) + y_(1)^(2)`
D. `x ^(2) + y^(2) +2x x _(1) + 2y y_(1) =0`

Answer

To find the equation of the circle that touches both axes and has its center at \((x_1, y_1)\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circle's Properties**:
- A circle that touches both the x-axis and y-axis means that the distance from the center of the circle to both axes is equal to the radius of the circle.
- If the center is at \((x_1, y_1)\), then the radius \(r\) of the circle is equal to both \(x_1\) and \(y_1\) because the circle must touch both axes.

2. **Set the Radius**:
- Since the circle touches both axes, we can conclude that:
\[ r = x_1 = y_1 \]

3. **Write the General Equation of a Circle**:
- The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

4. **Substitute the Values**:
- Here, substituting \(h = x_1\), \(k = y_1\), and \(r = x_1\) (or \(y_1\)), we get:
\[ (x - x_1)^2 + (y - y_1)^2 = x_1^2 \]

5. **Simplify the Equation**:
- Expanding the left side:
\[ (x - x_1)^2 + (y - y_1)^2 = x_1^2 \]
becomes:
\[ x^2 - 2x_1x + x_1^2 + y^2 - 2y_1y + y_1^2 = x_1^2 \]
- Now, subtract \(x_1^2\) from both sides:
\[ x^2 - 2x_1x + y^2 - 2y_1y + y_1^2 = 0 \]

6. **Combine Like Terms**:
- Since \(y_1 = x_1\), we can replace \(y_1\) with \(x_1\):
\[ x^2 - 2x_1x + y^2 - 2x_1y + x_1^2 = 0 \]

7. **Final Equation**:
- The final equation of the circle is:
\[ x^2 + y^2 - 2x_1(x + y) + x_1^2 = 0 \]