# What is the area of a circle given by the following equation: 2x^2 + 2y^2 + 4x + 4y = 0

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### 2 Answers

To find the area of the circle with the given equation `2x^2 + 2y^2 + 4x + 4y = 0` first find the radius of the circle.

First, simplify the equation by dividing every term by 2.

`x^2 + y^2 + 2x + 2y = 0`

Rewrite into general form of a circle `x^2 + y^2 = r^2` where r represents the radius of the circle.

To rewrite the equation, we must complete the square to find the radius.

`(x^2 + 2x + c) + (y^2 + 2y + d) = 0 + c + d`

To find c and d, divide middle term by 2 then square.

`(x^2 + 2x + 1) + (y^2 + 2y + 1) = 0 + 1 + 1`

`(x + 1)^2 + (y+1)^2 = 2`

This indicates that `r^2 = 2` so `r =sqrt(2)`

The radius of the circle is `sqrt(2).`

To find the area of a circle with radius `sqrt(2)` use the formula for area.

`A =pir^2`

`A =pi* (sqrt(2))^2`

`A = 2pi`

**The area of a circle given the equation `2x^2 + 2y^2 + 4x + 4y = 0` is `2pi.`**

The area of a circle represented by the equation `2x^2 + 2y^2 + 4x + 4y = 0` has to be determined.

`2x^2 + 2y^2 + 4x + 4y = 0`

=> `x^2 + y^2 + 2x + 2y = 0`

=> `x^2 + y^2 + 2x + 2y + 2 = 2`

=> `(x + 1)^2 + (y + 1)^2 = (sqrt 2)^2`

This is the equation of a circle with radius `sqrt 2` . The area of the circle is `pi*r^2 = pi*(sqrt 2)^2 = 2*pi`

**The area of the circle represented by the equation 2x^2 + 2y^2 + 4x + 4y = 0 is **`2*pi`