Let r0 = < x0, y0 >, r = <x, y>, and c > 0. Describe the set of all points P(x,y) such that ||r - r0|| = c.

Question 1:





Let r0 = %26lt; x0, y0 %26gt;, r = %26lt; x, y %26gt;, and c %26gt; 0. Describe the set of all points P(x,y) such that ||r - r0|| = c.





Question 2:





Let r0 = %26lt; x0, y0 %26gt;, r = %26lt; x, y %26gt;, and a = %26lt; a1, a2 %26gt; ≠ 0. Describe the set of all points P(x, y) such that r - r0 is a scalar multiple of a.








I think the answer for question 1 is a circle with center (x0, y0) and radius c.


I'm not really sure about question 2. r - r0 = %26lt; x-x0, y-y0 %26gt;, and there is a constant c such that %26lt; c(x-x0), c(y-y0) %26gt; = %26lt; a1, a2%26gt;. Should this also be a circle? Why? Should the constant c be the same for every x and y?

Let r0 = %26lt; x0, y0 %26gt;, r = %26lt;x, y%26gt;, and c %26gt; 0. Describe the set of all points P(x,y) such that ||r - r0|| = c.
yes,


for question 1 is a circle


namely, the circle (x-x0)^2+(y-y0)^2=c^2





for question, actually it's


r-r0=ca





so





r=r0+ca=(x0+ca1,y0+ca2)





Note that this is the vector equation of a line


the line that passes r0 and is parallel with a

jasmine