Circle Question?

Question

Find the center and radius for the circle defined by the equation x^2 + y^2 + 4x - 6y + 11 = 0

Answer 1

x^2+y^2+4x-6y+11=0
(x+2)^2+(y-3)^2+11-4-9=0
(x+2)^2(y-3)^2=sqrt(2)]^2
It is a circle with center (-2,3) and radius= sqrt(2)=1.4142.

Answer 2

The equation of the circle is, x^2 + y^2 + 4x - 6y + 11 = 0

now the general equation of circle is, x^2+y^2+2gx+2fy+c=0
and the radius = sq rt of (g^2+f^2-c) & center = (-g,-f)

so, if we compare these two equations we will find ,
2g = 4 or, g = 2 & 2f = -6 or, f = -3

so the center of the given circle will be ( -2 , 3) [ans1]

so the radius of the given circle will be sqrt of ( 2*2 + (-3)(-3) -11) i.e. sqrt of (4 + 9 - 11) i.e. sqrt of 2 i.e. 1.414..... [ans2]

Answer 3

Center(-2,3) radius sqrt(4+9-11)=sqrt(2)