Respuesta :

apologiabiology
complete the squaer for x's and y's seperately
actually, lets undsitribute the  2 for everybody first (except 14)

2(x^2-4x+y^2+6y)+14=0
group x and y's
2((x^2-4x)+(y^2+6y))+14=0
take 1/2 of the linear coefients and squaer them then add negative and positive of them inside
-4/2=-2, (-2)^2,  6/2=3, 3^2=9
2((x^2-4x+4-4)+(y^2+6y+9-9))+14=0
factor perfect squares
2(((x-2)^2-4)+((y+3)^2-9))+14=0
distribute
2(x-2)^2-8+2(y+3)^2-18+14=0
2(x-2)^2+2(y+3)^2-12=0
add 12 both sides
2(x-2)^2+2(y+3)^2=12
divide both sides by 2
(x-2)^2+(y+3)^2=6

for
(x-h)^2+(y-k)^2=r^2
center is (h,k)
radius is r


(x-2)^2+(y+3)^2=6
(x-2)^2+(y-(-3))^2=(√6)^2

center is (2,-3) and radius is√6

The equation of circle 2x² - 8x + 2y² + 12y + 14 = 0 center at (2, -3) and a radius is √6.

What is a circle?

It is a locus of a point drawn equidistant from the center. The distance from the center to the circumference is called the radius of the circle.

The given equation of the circle will be

[tex]2x^2-8x+2y^2+12y+14=0[/tex]

Divide the equation by 2, then we have

[tex]x^2-4x+y^2+6y+7=0[/tex]

Add 6 to both sides of the equation, we have

[tex]\begin{aligned} \rm x^2-4x+ 4 +y^2+6y+ 9 &=6\\\\\rm (x-2)^2 + (y+3)^2 &= (\sqrt{6})^2 \end{aligned}[/tex]

The center of the circle is (2, -3) and the radius of the circle is √6.

More about the circle link is given below.

https://brainly.com/question/11833983

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