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I want to find the equation but use vector length notation and I'm not sure about how to write it. $$ a) r = 2, A(-1; 1)$$

the line I'm not sure - $$|[x-x_0 , y-y_0]|^2 = r^2$$ then I do $$(x+1)^2 + (y-1)^2 = 4$$

Then for B) , the same thing but for a sphere $$ b) r = 9, A(3;-2;-1) $$

the line I'm not sure - $$|[x-x_0 , y-y_0, z-z0]|^2 = r^2$$ then I do $$(x-3)^2 + (y+2)^2 + (z+1)^2 = 81$$

Is this the right way to do it ? $$|[x-x_0 , y-y_0]|^2 = r^2$$

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1 Answer 1

up vote 1 down vote accepted

Let $p = (x,y)$ be arbitray point of the $xy$ plane. Circle is a locus of points that are equidistant from one point, called center of the circle. In this case it is $A(-1,1)$. To write this in vector notation: $$|p-A|=r$$ Now let's put the coordinates and the value for $r$: $$|(x+1,y-1)|= 2$$ I think this is what you are asked for. But you could put it also the other way: $$\sqrt{(x+1)^2+(y-1)^2}=2$$ or $$(x+1)^2 + (y-1)^2=4$$

Use which ever expretion is more sweatable for you.

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