The equation of the circle of radius `2sqrt(2)` whose centre lies on the line `x-y=0` and which touches the line `x+y=4`, and whose centre is coordinate satisfy `x+ygt4`, is

Options

A. `x^(2) +y^(2)-8x + 8y +24 =0`
B. ` x^(2) +y^(2) =8`
C. `x^(2) +y^(2)-8x+8y=24`
D. None of these

Answer

To solve the problem step by step, we need to find the equation of a circle with specific conditions. Let's break it down:

### Step 1: Understand the Given Information
- The radius of the circle \( r = 2\sqrt{2} \).
- The center of the circle lies on the line \( x - y = 0 \) (which means \( h = k \)).
- The circle touches the line \( x + y = 4 \).
- The center's coordinates \( (h, k) \) must satisfy \( h + k > 4 \).

### Step 2: Set Up the Circle's Equation
The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the radius:
\[ (x - h)^2 + (y - k)^2 = (2\sqrt{2})^2 = 8 \]

### Step 3: Substitute \( k \) in Terms of \( h \)
Since the center lies on the line \( x - y = 0 \), we have \( k = h \). Therefore, we can rewrite the equation as:
\[ (x - h)^2 + (y - h)^2 = 8 \]

### Step 4: Calculate the Distance from the Center to the Line
The distance \( d \) from the center \( (h, h) \) to the line \( x + y - 4 = 0 \) is given by the formula:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( x + y - 4 = 0 \), \( A = 1, B = 1, C = -4 \), and the center \( (h, h) \):
\[ d = \frac{|h + h - 4|}{\sqrt{1^2 + 1^2}} = \frac{|2h - 4|}{\sqrt{2}} \]
Since the circle touches the line, this distance must equal the radius:
\[ \frac{|2h - 4|}{\sqrt{2}} = 2\sqrt{2} \]

### Step 5: Solve for \( h \)
Multiplying both sides by \( \sqrt{2} \):
\[ |2h - 4| = 4 \]
This gives us two cases:
1. \( 2h - 4 = 4 \) → \( 2h = 8 \) → \( h = 4 \)
2. \( 2h - 4 = -4 \) → \( 2h = 0 \) → \( h = 0 \)

### Step 6: Determine Validity of \( h \)
Now, we need to check which value of \( h \) satisfies \( h + k > 4 \):
- For \( h = 4 \): \( h + k = 4 + 4 = 8 > 4 \) (valid)
- For \( h = 0 \): \( h + k = 0 + 0 = 0 \not> 4 \) (invalid)

Thus, the only valid center is \( (4, 4) \).

### Step 7: Write the Final Equation of the Circle
Substituting \( h = 4 \) and \( k = 4 \) into the circle's equation:
\[ (x - 4)^2 + (y - 4)^2 = 8 \]

### Step 8: Expand the Equation
Expanding the equation:
\[ (x^2 - 8x + 16) + (y^2 - 8y + 16) = 8 \]
Combining like terms:
\[ x^2 + y^2 - 8x - 8y + 32 - 8 = 0 \]
Thus, the final equation is:
\[ x^2 + y^2 - 8x - 8y + 24 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 8x - 8y + 24 = 0 \]