Question

Its distance from (0,0) is three times its distance from (4,0)

Answer 1

3,1

Answer 2

How?

Answer 3

see acc. to this x=3y
so if x=3
then y =1

Answer 4

How did you get that answer? Distance formula etc

Answer 5

id also suggest 3,0

Answer 6

\[x^2+y^2=(3d)^2\quad,\quad (x-4)^2+y^2=d^2\]\[x^2+y^2=9[(x-4)^2+y^2]\]\[x^2+y^2=9x^2-72x+144+9y^2\]\[0=8x^2-72x+144+8y^2\]\[0=x^2-9x+18+y^2\]\[0=x^2-9x+20.25-2.25+y^2\]\[1.5^2=(x-4.5)^2+y^2\]\[\mbox{all points on circle: }\quad 1.5^2=(x-4.5)^2+y^2\]

Answer 7

or, for the points (0,0) and (x0,y0),
all points on the circle
\[(x-\frac{9x_0}{8})^2+(y-\frac{9y_0}{8})^2 = \frac{9}{64}(x_0^2+y_0^2)\]
and for x0=4, y0=0, it reduces to the equation given by nikvist above.
Note that both (3,0) and (6,0) satisfy this question's requirements.