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

Question

king · Answer

3,1

anonymous · Answer

How?

king · Answer

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

anonymous · Answer

How did you get that answer? Distance formula etc

amistre64 · Answer

id also suggest 3,0

amistre64 · Answer

|dw:1326979014858:dw|

nikvist · Answer

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

mathmate · Answer

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.