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The given problem asks to find the point on the circle \((x - 5)^2 + (y - 4)^2 = 4\) which is closest to the other circle \((x - 1)^2 + (y - 1)^2 = 1\).

To solve this, follow these steps:

### Step 1: Identify the centers and radii of the circles.
- The first circle \((x - 5)^2 + (y - 4)^2 = 4\) has center \(C_1 = (5, 4)\) and radius \(r_1 = 2\) (since \(\sqrt{4} = 2\)).
- The second circle \((x - 1)^2 + (y - 1)^2 = 1\) has center \(C_2 = (1, 1)\) and radius \(r_2 = 1\) (since \(\sqrt{1} = 1\)).

### Step 2: Find the distance between the two centers.
The distance \(d\) between the centers \(C_1(5, 4)\) and \(C_2(1, 1)\) can be found using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substitute the coordinates of \(C_1\) and \(C_2\):
\[
d = \sqrt{(5 - 1)^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

### Step 3: Determine the closest point.
The closest point on the first circle to the second circle lies along the line between the centers. The direction vector from \(C_1\) to \(C_2\) is:
\[
\vec{v} = (1 - 5, 1 - 4) = (-4, -3)
\]
Normalize this vector to get the unit direction vector:
\[
\text{Magnitude of } \vec{v} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5
\]
\[
\hat{v} = \left( \frac{-4}{5}, \frac{-3}{5} \right)
\]
The closest point will be \(C_1\) shifted by a distance of \(r_1\) (radius of the first circle) in the direction of \(\hat{v}\). Thus, the point is:
\[
P = C_1 + r_1 \cdot \hat{v} = (5, 4) + 2 \cdot \left( \frac{-4}{5}, \frac{-3}{5} \right)
\]
\[
P = (5, 4) + \left( \frac{-8}{5}, \frac{-6}{5} \right) = \left( 5 + \frac{-8}{5}, 4 + \frac{-6}{5} \right)
\]
\[
P = \left( \frac{25}{5} + \frac{-8}{5}, \frac{20}{5} + \frac{-6}{5} \right) = \left( \frac{17}{5}, \frac{14}{5} \right)
\]

Thus, the point on the first circle closest to the second circle is \( \left( \frac{17}{5}, \frac{14}{5} \right) \).

Let me know if you need more clarification or further details! Here are 5 related questions you may find interesting:
1. How do you determine the point on a circle closest to a given point outside the circle?
2. How does the distance between the centers of two circles relate to the radius of the circles when they are externally tangent?
3. What is the geometric significance of the direction vector between two points in this problem?
4. How can you extend this approach to find the closest point on a circle to a line instead of another circle?
5. How do you solve problems involving tangents between two circles?

**Tip:** Always visualize the problem by sketching the circles, as it helps in understanding the relationship between the points and the circles.

Find the point on the circle (x - 5)^2 + (y - 4)^2 = 4 which is closest to the circle (x - 1)^2 + (y - 1)^2 = 1.

Solve the geometric problem of finding the point on the circle (x - 5)^2 + (y - 4)^2 = 4 that is closest to the circle (x - 1)^2 + (y - 1)^2 = 1. This solution covers the use of the distance formula and the geometric relationship between two circles.

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