A pair of straight lines passing through origin and the point of intersection of the curve `x^(2)+y^(2)=4` and the line `x+y=a`, are at right angle then the value of 'a' is

Options

A. 4
B. 2
C. `-4`
D. `-2`

Answer

To solve the problem, we need to find the value of 'a' such that the pair of straight lines passing through the origin and the point of intersection of the curve \(x^2 + y^2 = 4\) and the line \(x + y = a\) are at right angles.

### Step-by-Step Solution:

1. **Identify the Curve and Line**:
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This represents a circle centered at the origin (0,0) with a radius of 2.

2. **Find the Intersection Points**:
The line is given by:
\[ x + y = a \]
To find the points of intersection of the line and the circle, we can substitute \(y = a - x\) into the circle's equation:
\[ x^2 + (a - x)^2 = 4 \]
Expanding this gives:
\[ x^2 + (a^2 - 2ax + x^2) = 4 \]
\[ 2x^2 - 2ax + a^2 - 4 = 0 \]
This is a quadratic equation in \(x\).

3. **Use the Condition for Perpendicular Lines**:
For the lines through the origin and the point of intersection to be perpendicular, the product of their slopes must equal -1. The slope of the line \(x + y = a\) can be rearranged to:
\[ y = -x + a \]
Thus, the slope \(m_1 = -1\).

4. **Finding the Slopes of the Lines through the Origin**:
Let the slopes of the lines through the origin that intersect at the point \((x_0, y_0)\) be \(m_2\) and \(m_3\). The condition for perpendicularity gives:
\[ m_1 \cdot m_2 = -1 \]
Since \(m_1 = -1\), we have:
\[ (-1) \cdot m_2 = -1 \implies m_2 = 1 \]
Therefore, the lines through the origin will have slopes of 1 and -1.

5. **Finding the Intersection Points**:
The intersection points can be found by solving the quadratic equation:
\[ 2x^2 - 2ax + (a^2 - 4) = 0 \]
The discriminant of this quadratic must be non-negative for real intersection points:
\[ D = (-2a)^2 - 4 \cdot 2 \cdot (a^2 - 4) \geq 0 \]
Simplifying gives:
\[ 4a^2 - 8(a^2 - 4) \geq 0 \]
\[ 4a^2 - 8a^2 + 32 \geq 0 \]
\[ -4a^2 + 32 \geq 0 \implies 4a^2 \leq 32 \implies a^2 \leq 8 \]
Thus,
\[ -\sqrt{8} \leq a \leq \sqrt{8} \implies -2\sqrt{2} \leq a \leq 2\sqrt{2} \]

6. **Finding Specific Values of 'a'**:
Since the lines must pass through the origin and the intersection point, we can check specific values. The maximum distance from the origin to the circle is 2, so we can set \(a = 2\) and check if it satisfies the conditions.

7. **Conclusion**:
The value of \(a\) can be \(2\) or \(-2\) based on the perpendicularity condition and the intersection points being valid.

Thus, the final answer is:
\[ \boxed{2 \text{ or } -2} \]