# Is this right?

• May 26th 2010, 12:28 PM
iluvmathbutitshard
Is this right?
Question: If the equation x^2 + y^2 = 36 is changed to x^2 + (y+3)^2 = 36, what effect will this have on the graph.

I think that it will slide it 3 units up, but I'm not a hundred percent sure.
• May 26th 2010, 12:34 PM
Plato
Quote:

Originally Posted by iluvmathbutitshard
Question: If the equation x^2 + y^2 = 36 is changed to x^2 + (y+3)^2 = 36, what effect will this have on the graph.

I think that it will slide it 3 units up, but I'm not a hundred percent sure.

The center changes from $\displaystyle (0,0)$ to $\displaystyle (0,-3)$.
So?
• May 26th 2010, 12:38 PM
iluvmathbutitshard
So three units down?
• May 26th 2010, 01:46 PM
masters
Quote:

Originally Posted by iluvmathbutitshard
So three units down?

It would seem so.
• May 26th 2010, 03:07 PM
Quote:

Originally Posted by iluvmathbutitshard
Question: If the equation x^2 + y^2 = 36 is changed to x^2 + (y+3)^2 = 36, what effect will this have on the graph.

I think that it will slide it 3 units up, but I'm not a hundred percent sure.

Your own subconscious mind was correcting you.
You may luv maths but it's hard if the conscious does all the work.

Basically, the equation of a circle is Pythagoras' theorem applied to
all points on the circumference and the centre.

The distance formula is also Pythagoras' theorem

$\displaystyle \left(x-x_c\right)^2+\left(y-y_c\right)^2=r^2$

where $\displaystyle \left(x_c,\ y_c\right)$ are the x and y co-ordinates of the circle centre.

$\displaystyle x^2+(y+3)^2=36$

$\displaystyle [x-(0)]^2+[y-(-3)]^2=6^2$