Answer:
[tex]4) 2(x-3)^2+2(y+5)^2=96[/tex]
Step-by-step explanation:
[tex]2x^2 -12x +2y^2 + 20y - 28 = 0[/tex]
It basically represents a circle.
Equation of circle
[tex]\left(x-a\right)^2+\left(y-b\right)^2=r^2\:[/tex]
[tex]2x^2-12x+2y^2+20y=28[/tex]
[tex]$\frac{1}{2} (2x^2-12x+2y^2+20y)=28 \cdot \frac{1}{2} $[/tex]
[tex]x^2-6x+y^2+10y=14[/tex]
Organize the variables
[tex]\left(x^2-6x\right)+\left(y^2+10y\right)=14[/tex]
Now we have to complete the square for both groups:
[tex]\left(x^2-6x+9\right)+\left(y^2+10y+25\right)=14+9+25[/tex]
[tex]\left(x-3\right)^2+\left(y+5\right)^2=14+9+25[/tex]
[tex]\left(x-3\right)^2+\left(y+5\right)^2=48[/tex]
Now, I would proceed with:
[tex]\sqrt{48} =\sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3} =4\sqrt{3}[/tex]
[tex]\left(x-3\right)^2+\left(y+5\right)^2=(4\sqrt{3} )^2[/tex]
We got the center and the circle radius
[tex]\left(3,\:-5\right)\\r=4\sqrt{3}[/tex]
But this is not what you want:
So, just multiply both sides by 2 and you'll get:
[tex]2\left(x-3\right)^2+2\left(y+5\right)^2=96[/tex]
