Is this right?

**iluvmathbutitshard** · May 26th 2010, 12:28 PM

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.

**Plato** · May 26th 2010, 12:34 PM

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 $(0,0)$ to $(0,-3)$ .
So?

**iluvmathbutitshard** · May 26th 2010, 12:38 PM

So three units down?

**masters** · May 26th 2010, 01:46 PM

Originally Posted by iluvmathbutitshard

So three units down?

It would seem so.

**Archie Meade** · May 26th 2010, 03:07 PM

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

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

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

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

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

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