The point (2,n) lies on the circle whose equation is
(x-3)^2 + (y+2)^2=6. Find the value of n.

THANK YOU :)

Question

The point (2,n) lies on the circle whose equation is
 (x-3)^2 + (y+2)^2=6. Find the value of n. 

THANK YOU :)

anonymous · Answer

put $x=2$ get 
$$(2-3)^2+(y+2)^2=6$$ and solve for $y$

anonymous · Answer

Ok

anonymous · Answer

But we need to find n not y

anonymous · Answer

lol it is a number makes no difference what variable you use right?

anonymous · Answer

I guess

anonymous · Answer

if you are married to the $n$ put 
$$(2-3)^2+(n+2)^2=6$$ and solve for $n$

anonymous · Answer

Haha

anonymous · Answer

Ill try that

anonymous · Answer

ok
btw you should get two answers, because the circle will have to points on it with the first coordinate of $2$

anonymous · Answer

I got the square root of 5  and then subtract 2

anonymous · Answer

yes although it is really 
$$n+2=\pm\sqrt5$$

anonymous · Answer

because as i said there should be two answers

anonymous · Answer

So how exactly should I right the answer?

anonymous · Answer

Write*

anonymous · Answer

|dw:1425843825055:dw|

anonymous · Answer

$$n=-2\pm\sqrt5$$ is how i would write it

anonymous · Answer

Ohh, ok thanks :D

anonymous · Answer

Wait is it -5 or +5?

irishboy123 · Answer

Ashely, if you "got the square root of 5  and then subtract 2", there could be something amiss in your methodology and you might want to go through it again more slowly.

if it helps, post your work here in your own time and someone will be here to help if that is needed.