The equation of a circle which touches the y-axis at origin and whose radius is 3 units is

Options

A. `x^(2) + y^(2) pm 6y = 0 `
B. ` x^(2) + y^(2) pm 6x = 0`
C. `x^(2) + y^(2) pm 3 x = 0 `
D. ` x^(2) + y^(2) pm 3y = 0 `

Answer

To find the equation of a circle that touches the y-axis at the origin and has a radius of 3 units, we can follow these steps:

### Step 1: Understand the Position of the Circle
Since the circle touches the y-axis at the origin (0, 0) and has a radius of 3 units, the center of the circle must be 3 units away from the y-axis. Therefore, the center of the circle can either be at (3, 0) or (-3, 0).

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

### Step 3: Case 1 - Center at (3, 0)
If the center of the circle is at (3, 0):
- Here, \(h = 3\), \(k = 0\), and \(r = 3\).
- Substitute these values into the general equation:
\[ (x - 3)^2 + (y - 0)^2 = 3^2 \]
\[ (x - 3)^2 + y^2 = 9 \]

### Step 4: Expand the Equation
Now, expand the equation:
\[ (x - 3)^2 + y^2 = 9 \]
\[ x^2 - 6x + 9 + y^2 = 9 \]
Subtract 9 from both sides:
\[ x^2 - 6x + y^2 = 0 \]

### Step 5: Case 2 - Center at (-3, 0)
If the center of the circle is at (-3, 0):
- Here, \(h = -3\), \(k = 0\), and \(r = 3\).
- Substitute these values into the general equation:
\[ (x + 3)^2 + (y - 0)^2 = 3^2 \]
\[ (x + 3)^2 + y^2 = 9 \]

### Step 6: Expand the Equation
Now, expand the equation:
\[ (x + 3)^2 + y^2 = 9 \]
\[ x^2 + 6x + 9 + y^2 = 9 \]
Subtract 9 from both sides:
\[ x^2 + 6x + y^2 = 0 \]

### Step 7: Final Equations
Thus, the two possible equations of the circle are:
1. \(x^2 - 6x + y^2 = 0\) (for center at (3, 0))
2. \(x^2 + 6x + y^2 = 0\) (for center at (-3, 0))

### Conclusion
The final answer can be expressed as:
\[ x^2 + y^2 - 6x = 0 \quad \text{or} \quad x^2 + y^2 + 6x = 0 \]