Question

What are the coordinates of the center and the

Answer 1

Answer: The coordinate of the center is (4,-3)

radius = 8

Step-by-step explanation:

Given ;

$x^{2} +y^{2}=8x-6y+39$

re - arranging , we have

$x^{2} +y^{2}-8x+6y-39=0$

comparing with the general equation

$x^{2} +y^{2}+2gx+2fy+c=0$

This means that

$2gx = -8x$

$2g=-8$

$g=-4$

also

$2fy=6y$

$2f=6$

$f=3$

$c=-39$

The coordinate of the center is (-g,-f)

Therefore;

the coordinate of the center = (4,-3)

the formula for calculating r is given by:

$r = \sqrt{g^{2}+f^{2}-c}$

$r = \sqrt{16+9+39}$

$r=\sqrt{64}$

$r=8$

METHOD 2

$x^{2} +y^{2}-8x+6y-39=0$

using completing the square method , we have

$x^{2} -8x+y^{2}+6y-39=0$

completing the square , we have

$(x-4)^{2}+(y+3)^{2} = 39 +4^{2}+3^{2}$

$(x-4)^{2}+(y+3)^{2}=64$

comparing with the general equation

$(x-a)^{2}+(y-b)^{2}=r^{2}$

where a and b are the coordinate of the center , that is

the coordinate of the center is ( 4 , -3)

$r^{2}=64$

$r=8$