1. ## CIRCLE equation

The circle (x-h)^2+(y-3)^2=36 passes through the origins. Find the values of h. Would you substitute 0,0 for x and y?

2. Originally Posted by requal
The circle (x-h)^2+(y-3)^2=36 passes through the origins. Find the values of h. Would you substitute 0,0 for x and y?
Yes, that would be the first step.

3. yes, but that means I would get h^2+9=36 which becomes h^2=25 which therefore means that h=5,-5. However the answers says that it is $\displaystyle \pm3\sqrt{3}$

4. Originally Posted by requal
yes, but that means I would get h^2+9=36 which becomes h^2=25 which therefore means that h=5,-5. However the answers says that it is $\displaystyle \pm3\sqrt{3}$
In my universe 36 - 9 = 27 (most of the time, anyway).

5. eh, thanks anyway

6. Originally Posted by requal
eh, thanks anyway
I assume that means you understand where the correct answer has come from ....