Let's go through the problem step by step.

Step 1: Given Equation

The equation provided is:

$x^2 + y^2 = 1$

Step 2: Apply the Assumption Method (Let $x = 2$)

We are asked to assume $x = 2$ and then solve for $y$.

Substituting $x = 2$ into the equation:

$2^2 + y^2 = 1$

Simplifying further:

$4 + y^2 = 1$

Step 3: Solve for $y$

Now, solve for $y$:

$y^2 = 1 - 4$ $y^2 = -3$

Taking the square root of both sides:

$y = \pm\sqrt{-3}$

Since the square root of a negative number is not real, $y$ would be an imaginary number in this case:

$y = \pm i\sqrt{3}$

Step 4: Sketch the Graph

The original equation $x^2 + y^2 = 1$ represents a circle with a radius of 1 centered at the origin (0, 0).

However, substituting $x = 2$ into the equation leads to a value of $y$ that is not real, indicating that there are no real points on this circle corresponding to $x = 2$.

Step 5: Determine if $y$ is a Function of $x$

For $y$ to be a function of $x$, each $x$ value must correspond to exactly one $y$ value. However, the equation $x^2 + y^2 = 1$ typically corresponds to two values of $y$ for each $x$ within the domain $[-1, 1]$ (except at $x = \pm 1$).

Since $y = \pm \sqrt{1 - x^2}$, $y$ is not a function of $x$ because it does not pass the vertical line test—there are two values of $y$ for most values of $x$ in the domain.

Summary

• No real solution exists for $y$ when $x = 2$.
• The graph of $x^2 + y^2 = 1$ is a circle centered at the origin with a radius of 1.
• $y$ is not a function of $x$ because each $x$ does not map to a single $y$ value.

Would you like to see the detailed graph or have any other questions?

Here are 5 related questions to further explore the topic:

1. What would the graph of $x^2 + y^2 = 1$ look like if the radius were increased to 2?
2. How would the solution change if the assumption was $x = 0.5$?
3. What is the general form of a circle's equation in the coordinate plane?
4. Can you find the points where the circle intersects the x-axis and y-axis?
5. What does the vertical line test mean in terms of determining if a relation is a function?

Tip: When determining if a relation is a function, always check if each input (x-value) corresponds to only one output (y-value).