Question

Finding the center and radius by completing the square
4x(sqrd)+4y(sqrd)+4x-16y-19=0

Answer 1

`4x^2+4y^2+4x-16y-19=0 \\ 4x^2+4x+1+4y^2-16y+16-36=0\\ (2x+1)^2+(2y-4)^2=36`

center -
`(-1,4)\\`

`r=√(36)=6`

Answer 2

`(x-h)^2+(y-k)^2=r^2`
(h,k) - the center
r - the radius


`4x^2+4y^2+4x-16y-19=0 \ \ \ |/ 4 \\ x^2+y^2+x-4y-(19)/(4)=0 \\ x^2+x+(1)/(4)+y^2-4y+4-(19)/(4)-(1)/(4)-4=0 \\ (x+(1)/(2))^2+(y-2)^2-(20)/(4)-4=0 \\ (x+(1)/(2))^2+(y-2)^2-5-4=0 \\ (x+(1)/(2))^2+(y-2)^2-9=0 \\ (x+(1)/(2))^2+(y-2)^2=9 \\ (x+(1)/(2))^2+(y-2)^2=3^2`

The center -
`(-(1)/(2),2)`
The radius -
`3`