Equation: x² + y 2 = 1.
Solve the given equation using the “assumption method” let x = 1.  Sketch it after solving.Then determine whether y is a function of x.

### Step 1: Substitute \( x = 1 \) into the Equation
Given equation: 
\[
x^2 + y^2 = 1
\]

Substituting \( x = 1 \):
\[
1^2 + y^2 = 1
\]

This simplifies to:
\[
1 + y^2 = 1
\]

### Step 2: Solve for \( y \)
\[
y^2 = 1 - 1
\]
\[
y^2 = 0
\]
\[
y = 0
\]

So, when \( x = 1 \), \( y = 0 \).

### Step 3: Sketch the Solution
The equation \( x^2 + y^2 = 1 \) represents a circle with a radius of 1 centered at the origin \((0, 0)\) in the Cartesian plane.

The specific point we solved for is \((1, 0)\), which lies on the circle.

### Step 4: Determine Whether \( y \) is a Function of \( x \)
To determine if \( y \) is a function of \( x \), we check if each \( x \) value corresponds to only one \( y \) value. 

The original equation \( x^2 + y^2 = 1 \) represents a circle, and for most values of \( x \), there are two corresponding \( y \) values (one positive and one negative). For example:
- If \( x = 0 \), \( y^2 = 1 \), so \( y = 1 \) or \( y = -1 \).
- If \( x = \frac{1}{2} \), \( y^2 = \frac{3}{4} \), so \( y = \frac{\sqrt{3}}{2} \) or \( y = -\frac{\sqrt{3}}{2} \).

Since each \( x \) does not correspond to a unique \( y \), \( y \) is **not** a function of \( x \).

### Sketch and Visual Representation
To visualize this, imagine the circle centered at the origin, with points like \((1, 0)\), \((0, 1)\), \((-1, 0)\), and \((0, -1)\) lying on the circle. The point \((1, 0)\) is on the rightmost edge of the circle.

Would you like the sketch or any additional details on this solution?

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### Related Questions:
1. What is the general equation for a circle in a Cartesian plane?
2. Can \( y \) ever be a function of \( x \) for any equation of the form \( x^2 + y^2 = r^2 \)?
3. How would you solve the equation if the radius was larger, say \( r = 2 \)?
4. What happens to the function status of \( y \) if the equation was \( y = \sqrt{1 - x^2} \)?
5. How do you determine the domain and range for \( y \) in the equation \( x^2 + y^2 = 1 \)?

### Tip:
When checking if \( y \) is a function of \( x \), use the vertical line test: if any vertical line intersects the graph more than once, \( y \) is not a function of \( x \).

Explore how to solve the equation x² + y² = 1 using the assumption method. Understand why y is not a function of x for this equation and visualize the circle in the Cartesian plane. Suitable for advanced high school students.

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