Question 62662
Distance between P(x,y) and (0,-3) is given by:

D^2 = (x-0)^2 + (y-(-3))^2 

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

Likewise distance between P(x,y) and (0,3) is:

D1^2 = (x-0)^2 + (y-3)^2

     = x^2 + (y-3)^2

Given that the sum of the distances = 10

==> x^2 + (y+3)^2 + x^2 + (y-3)^2 = 100

==> x^2 + y^2 + 6y + 9 + x^2 + y^2 - 6y  + 9 = 100

==> 2x^2 + 2y^2 + 18 = 100

==> 2x^2 + 2y^2 = 100 - 18  [adding -18 to both the sides]

==> 2x^2 + 2y^2 = 82

==> x^2 + y^2 = 41 

This is the required equation.

Good Luck!!!