Please help me complete the square...I'm lost!
x^2+y^2-1/2=-x-y
To complete the square for the given equation x^2 + y^2 - 1/2 = -x - y, follow these steps:
Step 1: Group the x and y terms separately.
- Start by moving the x and y terms to one side of the equation:
x^2 + x + y^2 + y = -1/2
Step 2: Ensure the coefficient of x^2 and y^2 terms are both 1.
- In this case, the coefficients are already 1, so we can move on to the next step.
Step 3: Rearrange the equation.
- To complete the square, we need to rearrange the equation into a perfect square trinomial form.
- To do this, we need to halve the coefficients of x and y and square them:
x^2 + x + (1/2)^2 + y^2 + y + (1/2)^2 = -1/2 + (1/2)^2 + (1/2)^2
Step 4: Simplify the equation on both sides and write it in the perfect square trinomial form:
- On the left side, simplify the squared terms and combine the constants:
x^2 + x + 1/4 + y^2 + y + 1/4 = -1/2 + 1/4 + 1/4
- Simplify further:
x^2 + x + 1/4 + y^2 + y + 1/4 = 1/4
Step 5: Rewrite the equation as a perfect square trinomial.
- To do this, take the coefficient of the linear term (which is 1), divide it by 2, square the result, and add it to both sides:
x^2 + x + 1/4 + y^2 + y + 1/4 + 1/4 = 1/4 + 1/4
- Simplify further:
x^2 + x + 1/4 + y^2 + y + 1/4 = 1/2
- Now, rewrite the left side as a perfect square trinomial by factoring:
(x + 1/2)^2 + (y + 1/2)^2 = 1/2
So, the equation x^2 + y^2 - 1/2 = -x - y can be completed as:
(x + 1/2)^2 + (y + 1/2)^2 = 1/2
This is the equation of a circle centered at (-1/2, -1/2) with a radius of sqrt(1/2).