`x^2+y^2-10x-10y+41=0 and x^2+y^2-24x-10y+160=0` are two circle .Then the minimum distance between points lying on them is

Answer

To find the minimum distance between the points lying on the two given circles, we first need to determine the centers and radii of both circles from their equations.

### Step 1: Identify the first circle's parameters
The equation of the first circle is:
\[ x^2 + y^2 - 10x - 10y + 41 = 0 \]

We can rewrite this in the standard form of a circle:
\[ (x^2 - 10x) + (y^2 - 10y) + 41 = 0 \]

Completing the square for \(x\) and \(y\):
- For \(x^2 - 10x\):
\[ x^2 - 10x = (x - 5)^2 - 25 \]
- For \(y^2 - 10y\):
\[ y^2 - 10y = (y - 5)^2 - 25 \]

Substituting back, we get:
\[ (x - 5)^2 + (y - 5)^2 - 25 - 25 + 41 = 0 \]
\[ (x - 5)^2 + (y - 5)^2 - 9 = 0 \]
\[ (x - 5)^2 + (y - 5)^2 = 9 \]

Thus, the center of the first circle \(C_1\) is \((5, 5)\) and the radius \(r_1\) is:
\[ r_1 = \sqrt{9} = 3 \]

### Step 2: Identify the second circle's parameters
The equation of the second circle is:
\[ x^2 + y^2 - 24x - 10y + 160 = 0 \]

Rewriting in standard form:
\[ (x^2 - 24x) + (y^2 - 10y) + 160 = 0 \]

Completing the square for \(x\) and \(y\):
- For \(x^2 - 24x\):
\[ x^2 - 24x = (x - 12)^2 - 144 \]
- For \(y^2 - 10y\):
\[ y^2 - 10y = (y - 5)^2 - 25 \]

Substituting back, we get:
\[ (x - 12)^2 - 144 + (y - 5)^2 - 25 + 160 = 0 \]
\[ (x - 12)^2 + (y - 5)^2 - 9 = 0 \]
\[ (x - 12)^2 + (y - 5)^2 = 9 \]

Thus, the center of the second circle \(C_2\) is \((12, 5)\) and the radius \(r_2\) is:
\[ r_2 = \sqrt{9} = 3 \]

### Step 3: Calculate the distance between the centers
Now, we calculate the distance \(d\) between the centers \(C_1(5, 5)\) and \(C_2(12, 5)\):
\[ d = \sqrt{(12 - 5)^2 + (5 - 5)^2} = \sqrt{7^2 + 0^2} = \sqrt{49} = 7 \]

### Step 4: Calculate the minimum distance between the circles
The minimum distance \(D\) between the two circles is given by:
\[ D = d - (r_1 + r_2) \]
Substituting the values we found:
\[ D = 7 - (3 + 3) = 7 - 6 = 1 \]

### Final Answer
Thus, the minimum distance between points lying on the two circles is:
\[ \boxed{1} \]