Equation of the straight line which meets the circle `x^2 + y^2 = 8` at two points where these points are at a distance of 2 units from the point A(2, 2) is

Options

A. `x + y = 2`
B. `x + y = 3`
C. `x + y = 1`
D. `x + y = 0`

Answer

To find the equation of the straight line that meets the circle \( x^2 + y^2 = 8 \) at two points, where these points are at a distance of 2 units from the point \( A(2, 2) \), we can follow these steps:

### Step 1: Understand the Circle
The equation of the circle is given by:
\[ x^2 + y^2 = 8 \]
This can be rewritten as:
\[ x^2 + y^2 = (2\sqrt{2})^2 \]
This indicates that the circle has a radius of \( 2\sqrt{2} \) and is centered at the origin \( (0, 0) \).

### Step 2: Set Up the Distance Condition
We need to find points \( (x_1, y_1) \) and \( (x_2, y_2) \) on the circle such that the distance from point \( A(2, 2) \) to these points is 2 units. The distance formula gives us:
\[ \sqrt{(x_1 - 2)^2 + (y_1 - 2)^2} = 2 \]
Squaring both sides, we have:
\[ (x_1 - 2)^2 + (y_1 - 2)^2 = 4 \]

### Step 3: Expand the Distance Equation
Expanding the equation:
\[ (x_1^2 - 4x_1 + 4) + (y_1^2 - 4y_1 + 4) = 4 \]
This simplifies to:
\[ x_1^2 + y_1^2 - 4x_1 - 4y_1 + 8 = 4 \]
Rearranging gives:
\[ x_1^2 + y_1^2 - 4x_1 - 4y_1 + 4 = 0 \]

### Step 4: Substitute the Circle Equation
Since \( (x_1, y_1) \) lies on the circle, we know:
\[ x_1^2 + y_1^2 = 8 \]
Substituting this into the previous equation:
\[ 8 - 4x_1 - 4y_1 + 4 = 0 \]
This simplifies to:
\[ -4x_1 - 4y_1 + 12 = 0 \]
Dividing through by -4 gives:
\[ x_1 + y_1 = 3 \]

### Step 5: Equation of the Line
The equation \( x_1 + y_1 = 3 \) represents the straight line we are looking for. Thus, the equation of the line that meets the circle at the required points is:
\[ x + y = 3 \]

### Final Answer
The equation of the straight line is:
\[ \boxed{x + y = 3} \]