Question

What is the correct standard form of the equation

√(x+3)2+(y−2)2

explain each step

Answer 1

[AI-generated answer — Bot GPT 3.5]

The given equation is in the form of a circle centered at (-3, 2). To convert it into standard form, we need to expand and simplify the equation.

Step 1: Remove the square root.
√(x+3)^2 = |x+3|

Step 2: Square the expressions inside the absolute value and remove the parentheses.
|x+3|^2 = (x+3)^2 = x^2 + 6x + 9

Step 3: Apply the same procedure to the y-term.
|y-2|^2 = (y-2)^2 = y^2 - 4y + 4

Putting these two terms together, we have:
x^2 + 6x + 9 + y^2 - 4y + 4

Simplifying further, we combine like terms:
x^2 + 6x + y^2 - 4y + 13

Therefore, the correct standard form of the equation is:
x^2 + y^2 + 6x - 4y + 13 = 0

Answer 2

[AI-generated answer — Step-by-Step Bot]

To find the standard form of the equation √(x+3)² + (y-2)², we need to simplify and rewrite it without any radicals. Here are the steps:

Step 1: Expand and simplify the expression. Recall that (a+b)² = a² + 2ab + b².
(x+3)² + (y-2)² = x² + 6x + 9 + y² - 4y + 4

Step 2: Rearrange the terms so that all the variables are on the left side and the constants are on the right side.
x² + y² + 6x - 4y + 9 + 4 = 0

Step 3: Combine like terms and rearrange the equation in standard form.
x² + y² + 6x - 4y + 13 = 0

Therefore, the correct standard form of the equation √(x+3)² + (y-2)² is x² + y² + 6x - 4y + 13 = 0.