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To solve for \( y \) in the equation:

\[
2 \sqrt{5} = \sqrt{(5 - x)^2 + (4 - y)^2}
\]

we can start by squaring both sides to eliminate the square roots.

1. Square both sides:
   \[
   (2 \sqrt{5})^2 = \left(\sqrt{(5 - x)^2 + (4 - y)^2}\right)^2
   \]

   This simplifies to:
   \[
   4 \times 5 = (5 - x)^2 + (4 - y)^2
   \]
   \[
   20 = (5 - x)^2 + (4 - y)^2
   \]

2. Expand \((5 - x)^2\) and \((4 - y)^2\):
   \[
   20 = (25 - 10x + x^2) + (16 - 8y + y^2)
   \]

3. Combine like terms:
   \[
   20 = x^2 + y^2 - 10x - 8y + 41
   \]

4. Move 20 to the other side to simplify:
   \[
   0 = x^2 + y^2 - 10x - 8y + 21
   \]

This resulting equation represents a circle equation in terms of \(x\) and \(y\), but if you want to solve explicitly for \(y\), further steps would involve isolating \(y\). However, without additional values for \(x\), we might not be able to simplify it further directly for \(y\) alone.

If you have a specific value for \(x\) or want a general form, let me know so I can assist further.

Would you like a detailed breakdown of the steps so far, or do you have any questions?

### Related Questions:
1. How do we solve a similar equation for \(x\) if \(y\) is given?
2. Can we determine the radius and center of the circle from the final equation?
3. How would we graph this equation in a coordinate plane?
4. What happens if we substitute specific values for \(x\) and solve for \(y\)?
5. How can we check if a particular point \((x, y)\) satisfies this equation?

### Tip:
When dealing with square roots in equations, squaring both sides can help eliminate the roots, but remember it might introduce extraneous solutions.

Solve for y; 2√5 = √((5 - x)^2 + (4 - y)^2)

Learn how to solve for y in the equation 2√5 = √((5 - x)^2 + (4 - y)^2) by eliminating square roots and simplifying. This problem involves algebraic manipulation and the distance formula, suitable for high school students in Grades 10-12.

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