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

 Apr 21, 2020
 #1
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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

 Apr 21, 2020
 #2
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Guest used the right process but forgot that a "4" was placed on the right side of the equation.

 

                            2x2 - 8x + 2y2 + 12y + 14  =  0

Divide by 2:         x2 - 4x     + y2 + 6y     + 7  =  0

Subtract the 7:    x2 - 4x     + y2 + 6y            =  -7

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

                        ( x2 - 4x + 4 ) + ( y2 + 6y + 9 )  =  -7 + 4 + 9

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

 

Center:  (2, -3)

Radius:  sqrt(6)

 Apr 21, 2020

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