Question

x^2+y^2-4x-6y-12=0 in this circle find center and radius

Answer 1

Given eq. is 
x^2+y^2-4x-6y-12=0
x^2+y^2-4x-6y=12
Completing the squares within the parenthesis we get
x^2-4x+4+y^-6y+9=12+4+9
(x-2)^2+(y-3)^2=25 or (5)^2
Comparing given with the general eq. of the circle
(x-h)^2+(y-k)^2=r^2
we have
h=2, k=3 and r=5
so  center of the circle=(2,3) and radius of the circle is 5
hope it helps...

Answer 2

Answer:

(2,3) and r=5

Step-by-step explanation:

General equation of circle =

$x^{2}+y^{2}+2gx+2fy+c=0\\centre = (-g,-f)\\radius = \sqrt{g^{2}+f^{2} -c }$

on comparing:

$2g=-4\\g=-2\\2f=-6\\f=-3$

therefore centre = $(2,3)$

radius =

$\sqrt{2^{2} +3^{2}+12}\\=\sqrt{4+9+12}\\\= \sqrt{25} = 5$