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 :)

Answer 1

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

Answer 2

Ok

Answer 3

But we need to find n not y

Answer 4

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

Answer 5

I guess

Answer 6

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

Answer 7

Haha

Answer 8

Ill try that

Answer 9

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

Answer 10

I got the square root of 5 and then subtract 2

Answer 11

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

Answer 12

because as i said there should be two answers

Answer 13

So how exactly should I right the answer?

Answer 14

Write*

Answer 15

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

Answer 16

Ohh, ok thanks :D

Answer 17

Wait is it -5 or +5?

Answer 18

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.