Question

P= (x,y) is an arbitrary point on the circle x^2+y^2=36

Answer 1

Let (x,y) be an arbitrary point on the circle.

Then d = √[x-3)2 + (y-0)2] = √[x2 - 6x + 9 + y2]

Since x2 + y2 = 1, y2 = 1-x2.

So, d = √[x2 - 6x + 9 + 1 - x2]

     d = √(-6x+10)

Domain:  x is the x-coordinate of a point on the circle centered at (0,0) with radius 1.  So, -1≤x≤1.

             But, for d to be defined, we need -6x+10 ≥ 0.  So, x ≤ 5/3  (True for all x in [-1,1]).

             Domain = [-1,1]

Range:    -1 ≤ x ≤ 1   So, 6 ≥ -6x ≥ -6

                                    16 ≥ -6x+10 ≥ 4

                                     4 ≥ √(-6x+10) ≥ 2   That is, 2 ≤ d ≤4

              Range:  [2,4]

Step-by-step explanation: