Let's go through the steps to solve and analyze the equation:

Step 1: Given Equation

The equation provided is:

$x^2 + y^2 = 1$

Step 2: Apply the Assumption Method (Let $x = 4$)

We are asked to assume $x = 4$ and then solve for $y$.

Substituting $x = 4$ into the equation:

$4^2 + y^2 = 1$

Simplifying further:

$16 + y^2 = 1$

Step 3: Solve for $y$

Now, solve for $y$:

$y^2 = 1 - 16$ $y^2 = -15$

Taking the square root of both sides:

$y = \pm\sqrt{-15}$

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

$y = \pm i\sqrt{15}$

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 $x = 4$ 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 $x = 4$.

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.

In this case, since $x = 4$ gives no real solution for $y$, $y$ is not a function of $x$ over this domain.

Additionally, the equation $x^2 + y^2 = 1$ generally yields two values of $y$ for most $x$ values within the domain $[-1, 1]$, meaning it fails the vertical line test and is not a function of $x$.

Summary

• No real solution exists for $y$ when $x = 4$.
• 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 a detailed graph of the circle or have any other questions?

Here are 5 related questions to explore the topic further:

1. What is the maximum value of $x$ for which $y$ can have real solutions in the equation $x^2 + y^2 = 1$?
2. How would the graph change if the equation was $x^2 + y^2 = 4$?
3. What happens to the solutions for $y$ when $x$ is negative?
4. Can you find the points where the circle intersects the x-axis and y-axis?
5. How does the concept of imaginary numbers come into play when solving equations like this?

Tip: When working with the equation of a circle, remember that $x$ and $y$ values must satisfy $x^2 + y^2 \leq r^2$ for real number solutions, where $r$ is the radius of the circle.