Question

What is the center and radius?

X^2+y^2+6x-6y=-14

Answer 1

Therefore,

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

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=(6)/(2)=3\\ \\\therefore f=(-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=√((18-14)) \\\\\therefore radius = r=√(4)=2\ unit\\`

Therefore,

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