Just need an Ans. check please
a)
Find the eqation of the locus of a point which moves in the plane so that its distance from (-2,0) is always 3 times its distance from the origin.
my Ans: x^2+y^2+4x-32=0

b)
Show that the locus is a circle and state the center and radius of the circle.
My Ans.: (x+2)^2+(y-0)^2=36
r^2=36=6
center(-2,0)

Question

Just need an Ans. check please
a)
Find the eqation of the locus of a point which moves in the plane so that its distance from (-2,0) is always 3 times its distance from the origin.
my Ans: x^2+y^2+4x-32=0

b)
Show that the locus is a circle and state the center and radius of the circle.
My Ans.: (x+2)^2+(y-0)^2=36  
                                      r^2=36=6
                  center(-2,0)

anonymous · Answer

distance fron the point is always 6 - the curve is a circle radius 6 centre -2,0

( x + 2)^2 + y^2 = 36
x^2 + y^2 + 4x + 4 = 36
x^2 + y^2 + 4x + -32 = 0

you are correct

anonymous · Answer

sweet thnx

anonymous · Answer

since the curve is always the same ditsance from (-2,0) and this distance is 6, 6 is the radius of the circle and applying pythagoras
gives the first line in the last post

anonymous · Answer

welcome