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
What is the correct standard form of the equation
√(x+3)2+(y−2)2
explain each step
1 answer