A circle touches y-axis at (0, 2) and has an intercept of 4 units on the positive side of x-axis. The equation of the circle, is

Options

A. `x^(2)+y^(2)-4(sqrt(2)x+y)+4=0`
B. `x^(2)+y^(2)-4(x+sqrt(2)y)+4=0`
C. `x^(2)+y^(2)-2(sqrt(2)x+y)+4=0`
D. none of these

Answer

To find the equation of the circle that touches the y-axis at (0, 2) and has an intercept of 4 units on the positive side of the x-axis, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Point of Tangency and the Chord**:
- The circle touches the y-axis at the point \( P(0, 2) \).
- The circle has a chord on the x-axis that intercepts it at 4 units on the positive side. This means the chord extends from \( (0, 0) \) to \( (4, 0) \).

2. **Determine the Center of the Circle**:
- Since the circle touches the y-axis, the x-coordinate of the center will be equal to the radius \( R \).
- The y-coordinate of the center will be the same as the y-coordinate of the point of tangency, which is \( 2 \).
- Let the center be \( C(R, 2) \).

3. **Calculate the Radius**:
- The chord on the x-axis is 4 units long, meaning the distance from the center \( C \) to the chord is half of this length, which is \( 2 \) units.
- The distance from the center \( C \) to the x-axis (which is the y-coordinate of the center) is \( 2 \) units.
- We can use the Pythagorean theorem in the right triangle formed by the center \( C \), the midpoint of the chord \( D(2, 0) \), and the point of tangency \( P(0, 2) \).

\[ R^2 = (CD)^2 + (PD)^2 \]
where \( CD = 2 \) (the distance from the center to the x-axis) and \( PD = 2 \) (the half-length of the chord).

\[ R^2 = 2^2 + 2^2 = 4 + 4 = 8 \]
Thus, \( R = \sqrt{8} = 2\sqrt{2} \).

4. **Find the Coordinates of the Center**:
- The x-coordinate of the center is equal to the radius, so \( R = 2\sqrt{2} \).
- Therefore, the center \( C \) is at \( (2\sqrt{2}, 2) \).

5. **Write the Equation of the Circle**:
- The standard equation of a circle with center \( (h, k) \) and radius \( R \) is given by:
\[ (x - h)^2 + (y - k)^2 = R^2 \]
Substituting \( h = 2\sqrt{2} \), \( k = 2 \), and \( R^2 = 8 \):
\[ (x - 2\sqrt{2})^2 + (y - 2)^2 = 8 \]

6. **Expand the Equation**:
- Expanding the equation:
\[ (x - 2\sqrt{2})^2 = x^2 - 4\sqrt{2}x + 8 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
Combining these gives:
\[ x^2 - 4\sqrt{2}x + 8 + y^2 - 4y + 4 = 8 \]
Simplifying:
\[ x^2 + y^2 - 4\sqrt{2}x - 4y + 4 = 0 \]

### Final Equation of the Circle:
Thus, the equation of the circle is:
\[ x^2 + y^2 - 4\sqrt{2}x - 4y + 4 = 0 \]