Let the normals at all the points on a given curve pass through a fixed point (a,b). If the curve passes through (3,-3) and `(4,-2sqrt2)`, and given that `a-2sqrt2b=3`, then `(a^2+b^2+ab)` is equal to `"______"`

Answer

To solve the problem, we will follow these steps:

1. **Understanding the Problem**: We know that the normals at all points on a given curve pass through a fixed point \((a, b)\). The curve passes through the points \((3, -3)\) and \((4, -2\sqrt{2})\). We also have the condition \(a - 2\sqrt{2}b = 3\). We need to find the value of \(a^2 + b^2 + ab\).

2. **Assuming the Curve**: Since the normals at all points on the curve pass through a fixed point, the curve must be a circle with center at \((a, b)\).

3. **Finding the Radius**: The distance from the center \((a, b)\) to the points \((3, -3)\) and \((4, -2\sqrt{2})\) must be equal, as both points lie on the circle. We will use the distance formula to express this condition.

4. **Setting Up the Equations**:
- The distance from \((a, b)\) to \((3, -3)\) is given by:
\[ \sqrt{(a - 3)^2 + (b + 3)^2} \]
- The distance from \((a, b)\) to \((4, -2\sqrt{2})\) is given by:
\[ \sqrt{(a - 4)^2 + (b + 2\sqrt{2})^2} \]
- Setting these distances equal gives us:
\[ (a - 3)^2 + (b + 3)^2 = (a - 4)^2 + (b + 2\sqrt{2})^2 \]

5. **Expanding and Simplifying**:
- Expanding both sides:
\[ (a^2 - 6a + 9 + b^2 + 6b + 9) = (a^2 - 8a + 16 + b^2 + 4b + 8) \]
- This simplifies to:
\[ -6a + 18 + 6b = -8a + 24 + 4b \]
- Rearranging gives:
\[ 2a + 2b - 6 = 0 \quad \Rightarrow \quad a + b = 3 \]

6. **Using the Given Condition**: We have the equation \(a - 2\sqrt{2}b = 3\). Now we have two equations:
- \(a + b = 3\)
- \(a - 2\sqrt{2}b = 3\)

7. **Solving the System of Equations**:
- From \(a + b = 3\), we can express \(a\) as:
\[ a = 3 - b \]
- Substituting into the second equation:
\[ (3 - b) - 2\sqrt{2}b = 3 \]
\[ 3 - b - 2\sqrt{2}b = 3 \quad \Rightarrow \quad -b - 2\sqrt{2}b = 0 \quad \Rightarrow \quad b(1 + 2\sqrt{2}) = 0 \]
- Thus, \(b = 0\).

8. **Finding \(a\)**:
- Substituting \(b = 0\) into \(a + b = 3\):
\[ a + 0 = 3 \quad \Rightarrow \quad a = 3 \]

9. **Calculating \(a^2 + b^2 + ab\)**:
- Now we can find:
\[ a^2 + b^2 + ab = 3^2 + 0^2 + 3 \cdot 0 = 9 + 0 + 0 = 9 \]

Thus, the final answer is:
\[ \boxed{9} \]