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Show that the locus of a point, which moves so as always to be three times further from one fixed point than from another fixed point is a circle?

Habib
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Hint: We assume we are working in the plane. Let the two given fixed points be $(a,b)$ and $(c,d)$.

By the Pythagorean Theorem, the distance of a point (x,y) from (a,b) is $$\sqrt{(x-a)^2+(y-b)^2}.$$ The distance of $(x,y)$ from $(c,d)$ is $$\sqrt{(x-c)^2+(y-d)^2}.$$ The distance condition says that $$\sqrt{(x-a)^2+(y-b)^2}=3\sqrt{(x-c)^2+(y-d)^2}.$$ Square both sides, bring all terms to one side, divide by $8$, You will get a familiar kind of equation.

André Nicolas
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    You may as well shift the points so that $(a,b) = (0,0)$ and $(c,d) = (0,d)$. In other words, the first point is at the origin, and the second point is somewhere on the $x$-axis. Then you can do what Andre said, but the algebra will be a lot easier. – bubba Jul 29 '13 at 10:29
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    You're welcome. – missiledragon Jul 29 '13 at 11:14
  • I believe 'missile dragon' should be banned from maths stackexchange – Habib Jul 30 '13 at 09:18