robert7248
contestada

Using the method of completing the square, put each circle into the form
[tex](x-h)^{2} + (y-k)^{2} = r^{2}[/tex]
Then determine the center and radius of each circle

[tex]x^{2} + y^{2} +8x - 6y +16 = 0[/tex]

Respuesta :

kudzordzifrancis

ANSWER

Center; (-4,3)

Radius: r=3

EXPLANATION

The given circle has equation;

[tex]x^{2} + y^{2} +8x - 6y +16 = 0[/tex]

Rewrite to obtain;

[tex]x^{2} + y^{2} +8x - 6y = -16[/tex]

Regroup to obtain;

[tex]x^{2} +8x + y^{2}- 6y = -16[/tex]

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

[tex]x^{2} +8x + (4)^{2} + y^{2}- 6y + {( - 3)}^{2} = -16+ (4)^{2} + {( - 3)}^{2} [/tex]

[tex](x + 4)^{2} + {(y - 3)}^{2} = -16+ 16 + 9[/tex]

This simplifies to;

[tex](x + 4)^{2} + {(y - 3)}^{2} = {3}^{2} [/tex]

By comparison;

[tex](-4,3)=(h,k)[/tex]

and the radius is

[tex]r = 3[/tex]

Otras preguntas