The length of the tangent from any point on the circle `(x-3)^2 + (y + 2)^2 =5r^2` to the circle ` (x-3)^2 + (y + 2)^2 = r^2` is 4 units. Then the area between the circles is

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the circles and their properties
The equations of the circles are:
1. Circle S1: \((x-3)^2 + (y+2)^2 = 5r^2\)
2. Circle S2: \((x-3)^2 + (y+2)^2 = r^2\)

From these equations, we can determine:
- The center of both circles is at the point \(C(3, -2)\).
- The radius of Circle S1 is \(r_1 = \sqrt{5}r\).
- The radius of Circle S2 is \(r_2 = r\).

### Step 2: Use the tangent length information
We know that the length of the tangent from any point on Circle S1 to Circle S2 is given as 4 units.

Using the Pythagorean theorem in the triangle formed by the tangent, the radius of Circle S2, and the radius of Circle S1, we can write:
\[ CB^2 = AB^2 + CA^2 \]
Where:
- \(CB = r_1 = \sqrt{5}r\)
- \(AB = 4\) (length of the tangent)
- \(CA = r\)

Substituting the values:
\[ (\sqrt{5}r)^2 = 4^2 + r^2 \]
This simplifies to:
\[ 5r^2 = 16 + r^2 \]

### Step 3: Solve for \(r\)
Rearranging the equation gives:
\[ 5r^2 - r^2 = 16 \]
\[ 4r^2 = 16 \]
\[ r^2 = 4 \]
Taking the square root:
\[ r = 2 \]

### Step 4: Calculate the area between the circles
Now we need to find the area between the two circles:
\[ \text{Area between the circles} = \text{Area of Circle S1} - \text{Area of Circle S2} \]
The area of a circle is given by \(\pi r^2\).

Calculating the areas:
- Area of Circle S1:
\[ \text{Area of S1} = \pi (\sqrt{5}r)^2 = \pi (5r^2) = 5\pi r^2 \]
- Area of Circle S2:
\[ \text{Area of S2} = \pi r^2 \]

Substituting \(r^2 = 4\):
- Area of Circle S1:
\[ \text{Area of S1} = 5\pi (4) = 20\pi \]
- Area of Circle S2:
\[ \text{Area of S2} = \pi (4) = 4\pi \]

Now, finding the area between the circles:
\[ \text{Area between the circles} = 20\pi - 4\pi = 16\pi \]

### Final Answer
The area between the circles is \(16\pi\) square units.

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