The point (-a, - a) lies on the circle passing through the origin and the point of intersection of the straight line `x+y+a=0` with the circle `x^(2)+y^(2)=a^(2)`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while ensuring clarity and structure.

### Step 1: Understand the Problem
We need to find the equation of a circle that passes through the origin (0, 0) and the point of intersection of the line \(x + y + a = 0\) with the circle \(x^2 + y^2 = a^2\). The point \((-a, -a)\) is given to lie on this circle.

### Step 2: Find the Intersection Points
1. **Substitute the line equation into the circle equation**:
The line equation can be rewritten as \(y = -x - a\).
Substitute \(y\) into the circle equation:
\[ x^2 + (-x - a)^2 = a^2 \]
Expanding this:
\[ x^2 + (x^2 + 2ax + a^2) = a^2 \]
Simplifying:
\[ 2x^2 + 2ax + a^2 = a^2 \]
This simplifies to:
\[ 2x^2 + 2ax = 0 \]

### Step 3: Solve for \(x\)
2. **Factor the equation**:
\[ 2x(x + a) = 0 \]
This gives us two solutions:
\[ x = 0 \quad \text{or} \quad x = -a \]

### Step 4: Find Corresponding \(y\) Values
3. **Find \(y\) for each \(x\)**:
- If \(x = 0\):
\[ y = -0 - a = -a \quad \Rightarrow \quad (0, -a) \]
- If \(x = -a\):
\[ y = -(-a) - a = 0 \quad \Rightarrow \quad (-a, 0) \]

### Step 5: Identify Points of Intersection
The points of intersection are:
- \((0, -a)\)
- \((-a, 0)\)

### Step 6: Equation of the Circle
4. **Form the equation of the circle**:
The general form of the circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Since the circle passes through the origin (0, 0), substituting gives:
\[ 0 + 0 + 0 + 0 + c = 0 \quad \Rightarrow \quad c = 0 \]
So the equation simplifies to:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 7: Use Known Points to Find \(g\) and \(f\)
5. **Substituting the point \((-a, -a)\)**:
\[ (-a)^2 + (-a)^2 + 2g(-a) + 2f(-a) = 0 \]
This simplifies to:
\[ 2a^2 - 2ag - 2af = 0 \quad \Rightarrow \quad a^2 - ag - af = 0 \quad \Rightarrow \quad a^2 = a(g + f) \]
Thus:
\[ g + f = a \quad \text{(1)} \]

6. **Substituting the point \((0, -a)\)**:
\[ 0 + (-a)^2 + 2g(0) + 2f(-a) = 0 \]
This simplifies to:
\[ a^2 - 2af = 0 \quad \Rightarrow \quad a^2 = 2af \quad \Rightarrow \quad f = \frac{a}{2} \quad \text{(2)} \]

### Step 8: Find \(g\)
7. **Substituting \(f\) into equation (1)**:
\[ g + \frac{a}{2} = a \quad \Rightarrow \quad g = a - \frac{a}{2} = \frac{a}{2} \]

### Step 9: Final Equation of the Circle
8. **Substituting \(g\) and \(f\) back into the circle equation**:
\[ x^2 + y^2 + 2\left(\frac{a}{2}\right)x + 2\left(\frac{a}{2}\right)y = 0 \]
This simplifies to:
\[ x^2 + y^2 + ax + ay = 0 \]
Rearranging gives:
\[ x^2 + y^2 + ax + ay = 0 \]

### Final Answer
The equation of the required circle is:
\[ x^2 + y^2 + ax + ay = 0 \]