Question

What are the coordinates of the center and the
length of the radius of the circle whose equation is
x2 + y2 = 8x - 6y + 39?

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)

$r^{2}=64$

$r=8$

Answer 2

Completing the squares, it is found that:

• The coordinates of the center of the circle are (4,-3).

• The length of the radius is of 8 units.

What is the equation of a circle?

The equation of a circle of center $(x_0, y_0)$ and radius r is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

In this problem, the equation is given by:

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

Completing the squares, we have that:

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

$(x - 4)^2 + (y + 3)^2 = 39 + 16 + 9$

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

Hence:

• The coordinates of the center of the circle are (4,-3).

• The length of the radius is of 8 units.

More can be learned about the equation of a circle at https://brainly.com/question/24307696