Answer:
Center : (2,-3)
Radius : sqrt(6)
Step-by-step explanation:
Rewrite this is standard form to find the center and radius.
(x-2)^2 + (y+3)^2 = 6
From this, we can determine that the center is (2,-3) and the radius is sqrt(6)
Answer:
center is (2,-3)
Radius =[tex]\sqrt{6}[/tex]
Step-by-step explanation:
[tex]2x^2-8x+2y^2+12y+14=0[/tex]
To find out the center and radius we write the given equation in
(x-h)^2 +(y-k)^2 = r^2 form
Apply completing the square method
[tex]2x^2-8x+2y^2+12y+14=0[/tex]
[tex](2x^2-8x)+(2y^2+12y)+14=0[/tex]
factor out 2 from each group
[tex]2(x^2-4x)+2(y^2+6y)+14=0[/tex]
Take half of coefficient of middle term of each group and square it
add and subtract the numbers
4/2= 2, 2^2 = 4
6/2= 3, 3^2 = 9
[tex]2(x^2-4x+4-4)+2(y^2+6y+9-9)+14=0[/tex]
now multiply -4 and -9 with 2 to take out from parenthesis
[tex]2(x^2-4x+4)+2(y^2+6y+9)+14-8-18=0[/tex]
[tex]2(x-2)^2 +2(y+3)^2 -12=0[/tex]
Divide whole equation by 2
[tex](x-2)^2 +(y+3)^2 -6=0[/tex]
Add 6 on both sides
[tex](x-2)^2 +(y+3)^2 -6=0[/tex]
now compare with equation
(x-h)^2 + (y-k)^2 = r^2
center is (h,k) and radius is r
center is (2,-3)
r^2 = 6
Radius =[tex]\sqrt{6}[/tex]
