Question

Which equation is equivalent to the following?

2x^2 – 12x + 2y^2 + 20y – 28 = 0
1) 2(x-3)^2+2(y+5)^2=44
2) 2(x-3)^2+2(y+5)^2=60
3) 2(x-3)^2+2(y+5)^2=62
4) 2(x-3)^2+2(y+5)^2=96

Answer 1

`4) 2(x-3)^2+2(y+5)^2=96`

Explanation:

`2x^2 -12x +2y^2 + 20y - 28 = 0`

It basically represents a circle.

Equation of circle

`\left(x-a\right)^2+\left(y-b\right)^2=r^2\:`

`2x^2-12x+2y^2+20y=28`

`$(1)/(2) (2x^2-12x+2y^2+20y)=28 \cdot (1)/(2) $`

`x^2-6x+y^2+10y=14`

Organize the variables

`\left(x^2-6x\right)+\left(y^2+10y\right)=14`

Now we have to complete the square for both groups:

`\left(x^2-6x+9\right)+\left(y^2+10y+25\right)=14+9+25`

`\left(x-3\right)^2+\left(y+5\right)^2=14+9+25`

`\left(x-3\right)^2+\left(y+5\right)^2=48`

Now, I would proceed with:

`√(48) =√(2 \cdot 2 \cdot 2 \cdot 2 \cdot 3) =4√(3)`

`\left(x-3\right)^2+\left(y+5\right)^2=(4√(3) )^2`

We got the center and the circle radius

`\left(3,\:-5\right)\\r=4√(3)`

But this is not what you want:

So, just multiply both sides by 2 and you'll get:

`2\left(x-3\right)^2+2\left(y+5\right)^2=96`

Answer 2

4) 2(x - 3)^2 + 2(y + 5)^2 = 96.

Explanation:

2x^2 – 12x + 2y^2 + 20y – 28 = 0

2(x^2 – 6) + 2(y^2 + 10y) = 28

2 ( (x - 3)^2 - 9)) + 2((y + 5)^2 - 25) = 28

2(x - 3)^2 - 18 + 2(y + 5)^2 - 50 = 28

2(x - 3)^2 + 2(y + 5)^2 = 28 + 18 + 50

2(x - 3)^2 + 2(y + 5)^2 = 96