Question

Using the method of completing the square, put each circle into the form


`(x-h)^(2) + (y-k)^(2) = r^(2)`
Then determine the center and radius of each circle


`x^(2) + y^(2) +8x - 6y +16 = 0`

Answer 1

Center; (-4,3)

Radius: r=3

Step-by-step explanation

The given circle has equation;

`x^(2) + y^(2) +8x - 6y +16 = 0`

Rewrite to obtain;

`x^(2) + y^(2) +8x - 6y = -16`

Regroup to obtain;

`x^(2) +8x + y^(2)- 6y = -16`

Add the square of half the coefficient of the linear terms to both sides of the equation,

`x^(2) +8x + (4)^(2) + y^(2)- 6y + {( - 3)}^(2) = -16+ (4)^(2) + {( - 3)}^(2)`

`(x + 4)^(2) + {(y - 3)}^(2) = -16+ 16 + 9`

This simplifies to;

`(x + 4)^(2) + {(y - 3)}^(2) = {3}^(2)`

By comparison;

`(-4,3)=(h,k)`

and the radius is

`r = 3`