Question

What is the center and radius? X^2+y^2+6x-6y=-14

Answer 1

Answer:

Therefore,

$Center\ is\ C(-3,3)\\and\\Radius= 2\ units$

Step-by-step explanation:

Given:

A Equation of a Circle that is

$x^{2} +y^{2}+ 6x-6y=-14$

Which can be written as

$x^{2} +y^{2}+ 6x-6y+14=0$

To Find:

Center C( -g , -f ) = ?

radius = r =?

Solution:

General Equation of a Circle is given as

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

On Comparing the Given equation with the General equation we get the following values,

$2g=6\ and\ 2f=-6\ and\ c=14\\\therefore g=\frac{6}{2}=3\\ \\\therefore f=\frac{-6}{2}=-3$

Now, Center and Radius for above circle is given as

$C(-g,-f)=(-3,-(-3))=(-3,3)\\and\\radius = r=\sqrt{(g^{2}+f^{2}-c )} = \sqrt{(3^{2}+(-3)^{2}-14)}\\\\\therefore radius = r=\sqrt{(18-14)} \\\\\therefore radius = r=\sqrt{4}=2\ unit$

Therefore,

$Center\ is\ C(-3,3)\\and\\Radius= 2\ units$