If a chord of circle `x^2 + y^2 =8` makes equal intercepts of length ‘a’ on the coordinate axes then `|a| <`

Options

A. `|a| lt 2`
B. ` |a| lt sqrt(2)`
C. ` |a| lt 4`
D. `|a| gt 4`

Answer

To solve the problem, we need to determine the condition on the length of the intercepts made by a chord of the circle defined by the equation \(x^2 + y^2 = 8\).

### Step-by-Step Solution:

1. **Identify the Circle's Properties**:
The given equation of the circle is:
\[ x^2 + y^2 = 8 \]
The center of the circle is at the origin \((0, 0)\) and the radius \(r\) can be calculated as:
\[ r = \sqrt{8} = 2\sqrt{2} \]

2. **Equation of the Chord**:
The chord makes equal intercepts of length \(a\) on the coordinate axes. Therefore, the intercepts on the x-axis and y-axis are both \(a\). The equation of the line making these intercepts can be expressed as:
\[ \frac{x}{a} + \frac{y}{a} = 1 \quad \text{or} \quad x + y = a \]

3. **Distance from the Origin to the Chord**:
The distance \(d\) from the origin \((0, 0)\) to the line \(x + y = a\) can be calculated using the formula for the distance from a point to a line:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 1\), \(B = 1\), and \(C = -a\). Thus, the distance becomes:
\[ d = \frac{|0 + 0 - a|}{\sqrt{1^2 + 1^2}} = \frac{|a|}{\sqrt{2}} \]

4. **Condition for the Chord**:
Since the chord lies inside the circle, the distance from the origin to the chord must be less than the radius of the circle:
\[ \frac{|a|}{\sqrt{2}} < 2\sqrt{2} \]

5. **Simplifying the Inequality**:
Multiply both sides of the inequality by \(\sqrt{2}\):
\[ |a| < 2\sqrt{2} \cdot \sqrt{2} \]
This simplifies to:
\[ |a| < 2 \cdot 2 = 4 \]

### Final Result:
Thus, we conclude that:
\[ |a| < 4 \]