Show that the circle `x^(2)+ y^(2) - 4x + 4y + 4 = 0` touches the co-ordinate axes. If the points of contact are A and B, find the equation of the circle which passes through A, B and the origin,

Answer

To solve the problem step by step, we will first rewrite the equation of the circle in standard form, identify its center and radius, and then check if it touches the coordinate axes. Finally, we will find the equation of the circle that passes through the points of contact and the origin.

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 4x + 4y + 4 = 0 \]

We can rearrange this equation to make it easier to analyze:
\[ x^2 - 4x + y^2 + 4y + 4 = 0 \]

### Step 2: Complete the square
To rewrite the equation in standard form, we will complete the square for both \(x\) and \(y\).

1. For \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

2. For \(y^2 + 4y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

Now substituting back into the equation:
\[ (x - 2)^2 - 4 + (y + 2)^2 - 4 + 4 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y + 2)^2 - 4 = 0 \]
Thus, we have:
\[ (x - 2)^2 + (y + 2)^2 = 4 \]

### Step 3: Identify the center and radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \(C(2, -2)\)
- Radius \(r = 2\)

### Step 4: Check if the circle touches the coordinate axes
To check if the circle touches the coordinate axes, we need to see if the distance from the center to each axis is equal to the radius.

1. **Distance to the x-axis**: The distance from the center \(C(2, -2)\) to the x-axis (y=0) is \(|-2| = 2\).
2. **Distance to the y-axis**: The distance from the center \(C(2, -2)\) to the y-axis (x=0) is \(|2| = 2\).

Since both distances are equal to the radius, the circle touches both axes.

### Step 5: Find the points of contact
- The point of contact with the x-axis (A) is \((2, 0)\).
- The point of contact with the y-axis (B) is \((0, -2)\).

### Step 6: Find the equation of the circle passing through A, B, and the origin
The points we have are:
- A: \((2, 0)\)
- B: \((0, -2)\)
- Origin: \((0, 0)\)

The general equation of a circle can be written as:
\[ x^2 + y^2 - 2gx - 2fy + c = 0 \]

### Step 7: Substitute the points into the circle equation
1. For the origin \((0, 0)\):
\[ 0^2 + 0^2 - 2g(0) - 2f(0) + c = 0 \]
This gives us \(c = 0\).

2. For point A \((2, 0)\):
\[ 2^2 + 0^2 - 2g(2) - 2f(0) + 0 = 0 \]
\[ 4 - 4g = 0 \]
Thus, \(g = 1\).

3. For point B \((0, -2)\):
\[ 0^2 + (-2)^2 - 2g(0) - 2f(-2) + 0 = 0 \]
\[ 4 + 4f = 0 \]
Thus, \(f = -1\).

### Step 8: Write the final equation of the circle
Substituting \(g\), \(f\), and \(c\) back into the general equation:
\[ x^2 + y^2 - 2(1)x - 2(-1)y + 0 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x + 2y = 0 \]

### Final Answer:
The equation of the circle that passes through points A, B, and the origin is:
\[ x^2 + y^2 - 2x + 2y = 0 \]