What is the center and radius for the circle with equation 2x^2 -8x + 2y^2 +12y +14 = 0

Guest Apr 21, 2020

#1**+1 **

we can use difference of squares.

first, let's divide everything on both sides of the equation by 2... then, we have:

x^2 - 4x + y^2 + 6y + 7 = 0

Now, we add 4 on both sides in order to have (x^2 - 4x + 4) + y^2 + 6y + 7 = 4, meaning we can rewrite (x^2 - 4x + 4) as (x-2)^2.

Again, we can do the same...

we subtract 7 on both sides to get (x-2)^2 + y^2 + 6y = -7...

Seeing the y^2 + 6y, we can also think of y^2 + 6y + 9 which allows us to rewrite this as (y+3)^2...

So, we add 9 to both sides to get (x-2)^2 + (y+3)^2 = 2...

Now, this equation is in the circle form (x-h)^2 + (y-k)^2 = r^2 where r is the radius and (h,k) is the center of the circle.

2 = r^2

r = sqrt2

So, this circle has center (2,-3) and has radius sqrt2.

hope this helped! :D

Guest Apr 21, 2020

#2**0 **

Guest used the right process but forgot that a "4" was placed on the right side of the equation.

2x^{2} - 8x + 2y^{2} + 12y + 14 = 0

Divide by 2: x^{2} - 4x + y^{2} + 6y + 7 = 0

Subtract the 7: x^{2} - 4x + y^{2} + 6y = -7

Complete the squares by adding 4 and 9 to both sides:

( x^{2} - 4x + 4 ) + ( y^{2} + 6y + 9 ) = -7 + 4 + 9

Rewrite: (x - 2)^{2} + (y + 3)^{2} = 6

Center: (2, -3)

Radius: sqrt(6)

geno3141 Apr 21, 2020