a circle passing through the point `(2,2(sqrt2-1))` touches the pair of lines `x^2 -y^2 -4x+4 =0`. The centre of the circle is

Options

A. `(2,2sqrt(2))` and `(2,6sqrt(6)-8)`
B. `(2,5sqrt(2))` and `(2,7sqrt(2))`
C. `(2,5sqrt(2)-1)` and `(2,-3)`
D. None of these

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the given information
We have a circle that passes through the point \( (2, 2(\sqrt{2} - 1)) \) and touches the pair of lines given by the equation \( x^2 - y^2 - 4x + 4 = 0 \).

### Step 2: Rewrite the equation of the lines
The equation \( x^2 - y^2 - 4x + 4 = 0 \) can be factored as follows:
\[ x^2 - 4x + 4 - y^2 = 0 \implies (x - 2)^2 - y^2 = 0 \]
This can be rewritten as:
\[ (x - 2)^2 = y^2 \]
Taking the square root gives us the two lines:
\[ x - y = 2 \quad \text{and} \quad x + y = 2 \]

### Step 3: Identify the center of the circle
Since the circle touches both lines, its center must lie on the angle bisector of these two lines. The intersection of the lines occurs at the point \( (2, 0) \). We can denote the center of the circle as \( (2, k) \).

### Step 4: Use the distance formula
The distance from the center \( (2, k) \) to one of the lines (let's use \( x - y = 2 \)) must equal the radius of the circle. The distance \( d \) from a point \( (p, q) \) to a line \( ax + by + c = 0 \) is given by:
\[ d = \frac{|ap + bq + c|}{\sqrt{a^2 + b^2}} \]
For the line \( x - y - 2 = 0 \) (where \( a = 1, b = -1, c = -2 \)), the distance from the point \( (2, k) \) is:
\[ d = \frac{|1 \cdot 2 - 1 \cdot k - 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|2 - k - 2|}{\sqrt{2}} = \frac{| -k |}{\sqrt{2}} = \frac{k}{\sqrt{2}} \]

### Step 5: Set the distance equal to the radius
The radius of the circle is the distance from the center \( (2, k) \) to the point \( (2, 2(\sqrt{2} - 1)) \):
\[ \text{Radius} = |k - 2(\sqrt{2} - 1)| \]
Setting the two distances equal gives:
\[ \frac{k}{\sqrt{2}} = |k - 2(\sqrt{2} - 1)| \]

### Step 6: Solve the equation
We will solve this equation in two cases (since we have the absolute value).

**Case 1:** \( k - 2(\sqrt{2} - 1) \geq 0 \)
\[ \frac{k}{\sqrt{2}} = k - 2\sqrt{2} + 2 \]
Multiplying through by \( \sqrt{2} \):
\[ k = k\sqrt{2} - 2\sqrt{2} + 2\sqrt{2} \]
Rearranging gives:
\[ k(\sqrt{2} - 1) = 2\sqrt{2} \]
Thus,
\[ k = \frac{2\sqrt{2}}{\sqrt{2} - 1} \]

**Case 2:** \( k - 2(\sqrt{2} - 1) < 0 \)
\[ \frac{k}{\sqrt{2}} = -k + 2\sqrt{2} - 2 \]
Multiplying through by \( \sqrt{2} \):
\[ k = -k\sqrt{2} + 2\cdot2 - 2\sqrt{2} \]
Rearranging gives:
\[ k(1 + \sqrt{2}) = 4 - 2\sqrt{2} \]
Thus,
\[ k = \frac{4 - 2\sqrt{2}}{1 + \sqrt{2}} \]

### Step 7: Simplify the results
Calculating both values of \( k \):
1. From Case 1:
\[ k = 2\sqrt{2} + 2 \]
2. From Case 2:
\[ k = 2\sqrt{2} - 6 \]

### Conclusion
The two possible centers of the circle are:
1. \( (2, 2\sqrt{2} + 2) \)
2. \( (2, 2\sqrt{2} - 6) \)