Let $a,\ b$ be real constants. If real numbers $x,\ y$ satisfy $x^2+y^2\leq 25,\ \ 2x+y\leq 5$, then find the minimum value of 
$z=x^2+y^2-2ax-2by$.
To find the minimum value of \( z = x^2 + y^2 - 2ax - 2by \) under the constraints \( x^2 + y^2 \leq 25 \) and \( 2x + y \leq 5 \), we can proceed as follows:

1. **Rewrite the expression for \( z \):**
   \[
   z = x^2 + y^2 - 2ax - 2by
   \]
   We can complete the square for \( x \) and \( y \):
   \[
   z = (x^2 - 2ax + a^2) + (y^2 - 2by + b^2) - a^2 - b^2
   \]
   \[
   z = (x - a)^2 + (y - b)^2 - a^2 - b^2
   \]

2. **Interpret the expression geometrically:**
   The term \((x - a)^2 + (y - b)^2\) represents the squared distance from the point \((x, y)\) to the point \((a, b)\). Therefore, we can rewrite \( z \) as:
   \[
   z = \text{distance}^2 - a^2 - b^2
   \]
   where \(\text{distance} = \sqrt{(x - a)^2 + (y - b)^2}\).

3. **Analyze the constraints:**
   - The constraint \( x^2 + y^2 \leq 25 \) represents a circle of radius 5 centered at the origin.
   - The constraint \( 2x + y \leq 5 \) represents a half-plane.

4. **Determine the minimum distance:**
   - If the point \((a, b)\) lies inside the region defined by the constraints, the minimum distance from \((a, b)\) to any point \((x, y)\) in the region is 0.
   - If the point \((a, b)\) lies outside the region, we need to find the point on the boundary of the region that is closest to \((a, b)\).

5. **Case 1: \( (a, b) \) inside the region:**
   If \((a, b)\) is inside the region, the minimum distance is 0. Therefore:
   \[
   z = 0 - a^2 - b^2 = -a^2 - b^2
   \]

6. **Case 2: \( (a, b) \) outside the region:**
   If \((a, b)\) is outside the region, we need to find the closest point on the boundary. The boundary consists of the circle \( x^2 + y^2 = 25 \) and the line \( 2x + y = 5 \).

   - For the circle \( x^2 + y^2 = 25 \):
     The closest point on the circle to \((a, b)\) can be found by projecting \((a, b)\) onto the circle. The distance from \((a, b)\) to the origin is \(\sqrt{a^2 + b^2}\). If this distance is greater than 5, the closest point on the circle is:
     \[
     \left( \frac{5a}{\sqrt{a^2 + b^2}}, \frac{5b}{\sqrt{a^2 + b^2}} \right)
     \]
     The distance from \((a, b)\) to this point is:
     \[
     \sqrt{ \left( a - \frac{5a}{\sqrt{a^2 + b^2}} \right)^2 + \left( b - \frac{5b}{\sqrt{a^2 + b^2}} \right)^2 }
     \]
     Simplifying, this distance is:
     \[
     \sqrt{a^2 + b^2} - 5
     \]

   - For the line \( 2x + y = 5 \):
     The closest point on the line to \((a, b)\) can be found using the formula for the distance from a point to a line. The distance is:
     \[
     \frac{|2a + b - 5|}{\sqrt{2^2 + 1^2}} = \frac{|2a + b - 5|}{\sqrt{5}}
     \]

   We need to compare these distances and choose the smaller one.

7. **Conclusion:**
   The minimum value of \( z \) is achieved when the distance is minimized. If \((a, b)\) is inside the region, the minimum value is \(-a^2 - b^2\). If \((a, b)\) is outside the region, we need to find the minimum distance to the boundary and use that to compute \( z \).

The final answer is \( \boxed{ -a^2 - b^2 } \) if \((a, b)\) is inside the region, otherwise it depends on the minimum distance to the boundary.