In a square with a side of 2 units, a circle with the center of one of its vertices and a radius of 2.5 units intersects the two sides of the square, what is the distance from the nearest vertex of the square to the point of intersection?

Let's solve this problem step by step.

### Step 1: Understanding the Geometry
- You have a square with a side length of 2 units.
- A circle is centered at one of the vertices of the square, with a radius of 2.5 units.
- The circle intersects the two sides of the square.

### Step 2: Setting Up the Coordinate System
Place the square in a coordinate system where one vertex of the square (which is also the center of the circle) is at the origin \((0, 0)\). The vertices of the square are then:
- \((0, 0)\)
- \((2, 0)\)
- \((2, 2)\)
- \((0, 2)\)

The equation of the circle with center at \((0, 0)\) and radius \(2.5\) is:
\[
x^2 + y^2 = 2.5^2 = 6.25
\]

### Step 3: Finding Points of Intersection
To find the points where the circle intersects the sides of the square, we can use the equations of the lines along the sides of the square:
1. For the side along the \(x\)-axis, \(y = 0\).
2. For the side along the \(y\)-axis, \(x = 0\).

#### Intersection with the \(x\)-axis (\(y = 0\)):
Substituting \(y = 0\) into the circle's equation:
\[
x^2 + 0^2 = 6.25
\]
\[
x^2 = 6.25 \quad \Rightarrow \quad x = \sqrt{6.25} = 2.5
\]
So, the point of intersection on the \(x\)-axis is \((2.5, 0)\).

#### Intersection with the \(y\)-axis (\(x = 0\)):
Substituting \(x = 0\) into the circle's equation:
\[
0^2 + y^2 = 6.25
\]
\[
y^2 = 6.25 \quad \Rightarrow \quad y = \sqrt{6.25} = 2.5
\]
So, the point of intersection on the \(y\)-axis is \((0, 2.5)\).

### Step 4: Distance from the Nearest Vertex
The nearest vertex to both points \((2.5, 0)\) and \((0, 2.5)\) is the origin \((0, 0)\). The distance from \((0, 0)\) to either point is the same and can be calculated using the distance formula:
\[
\text{Distance} = \sqrt{(2.5 - 0)^2 + (0 - 0)^2} = \sqrt{2.5^2 + 0^2} = \sqrt{6.25} = 2.5
\]

### Conclusion
The distance from the nearest vertex of the square (which is the origin in this case) to the point of intersection is **2.5 units**.

Would you like more details or have any other questions?

Here are 8 related questions:
1. How do you find the equation of a circle in a different coordinate system?
2. What happens when the circle has a radius less than the side length of the square?
3. How would the problem change if the circle's radius was exactly 2 units?
4. Can you calculate the area of the circle that lies within the square?
5. What is the distance from the center of the square to the point of intersection?
6. How does the equation of a circle change if its center is not at the origin?
7. Can you derive the equation of the square's diagonal line and its intersections with the circle?
8. What is the significance of the Pythagorean Theorem in this problem?

**Tip:** Always visualize problems involving geometry and coordinates by sketching a diagram; it makes understanding the relationships easier.

Explore how to calculate the distance from the nearest vertex of a square to the point where a circle with a radius of 2.5 units intersects the square. This problem involves coordinate geometry and the Pythagorean Theorem. Suitable for high school students.

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