I just verify my solution for a problem below:
Obtain the centre and radius of the circle
x^2 + y^2 — 6x —12y —19 = 0.

Can anybody help me?

Question

I just verify my solution for a problem below:
Obtain the centre and radius of the circle
x^2 + y^2 — 6x —12y —19 = 0.

Can anybody help me?

anonymous · Answer

Proceed by completing the square. From there, what did you get and I can verify it :P

anonymous · Answer

My answer is radius : 8 and cneter is at (3,6)

anonymous · Answer

Okay, just 2 seconds :P

jit4won · Answer

center = 6/2 . 12/2= 3,6

jit4won · Answer

radius=8

anonymous · Answer

I got:
$$(x-3)^2+(y-6)^2=64$$.
So yes, radius 8 and center (3,6)

anonymous · Answer

Thanks for your help.I appreciate it.

anonymous · Answer

No problem :P

jit4won · Answer

if you have an equation in the form of x^2 + Y^2 +2gx +2fy+c= 0 use 
center = -g,-f

anonymous · Answer

Oh,that's new for me.Thanks jit4won.

jit4won · Answer

radius = square root of   ( g^2+f^2 - c)

jit4won · Answer

Its a shortcut method ..

jit4won · Answer

please check it now

jit4won · Answer

center =( - half of coefficient of x,  - half of coefficient of y)

jit4won · Answer

I can help you if you have skype/gtalk /yahoo/ any other site where i could instruct you