Hello, xwrathbringerx!
Looks like you need a walkthrough . . . hints don't seem to work.
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.
Since this claim is made for any two points,
. . place the points on the xaxis, symmetric to the origin. Code:
 P
 o (x,y)
 * *
*  *
*  *
  o     +     o  
(a,0)  (a,0)
A  B

Let $\displaystyle P(x,y)$ be any point such that: .$\displaystyle PA \:=\:3\!\cdot\!PB$
Using the Distance Formula: .$\displaystyle \begin{array}{ccc}PA &=& \sqrt{(x+a)^2 + y^2} \\ \\[3mm] PB &=& \sqrt{(xa)^2 + y^2} \end{array}$
Then: .$\displaystyle PA \:=\:3\!\cdot\!PB \quad\Rightarrow\quad \sqrt{(x+a)^2 + y^2} \:=\:3\sqrt{(xa)^2 + y^2} $
Square both sides: .$\displaystyle (x+a)^2 + y^2 \:=\:9\left[(xa)^2 + y^2\right]$
Expand: .$\displaystyle x^2 + 2ax + a^2 + y^2 \:=\:9x^2  18ax + 9a^2 + 9y^2 $
Simplify: .$\displaystyle 8x^2  20ax + 8y^2 \:=\:8a^2$
Divide by 8: .$\displaystyle x^2  \tfrac{5}{2}ax + h^2 \:=\:a^2$
Complete the square: .$\displaystyle x^2  \tfrac{5}{2}ax + {\color{blue}\tfrac{25}{16}a^2} \:=\:a^2 + {\color{blue}\tfrac{25}{16}a^2}$
. . And we have: .$\displaystyle \left(x\tfrac{5}{4}a\right)^2 + y^2 \:=\:\tfrac{9}{16}a^2$
This is a circle with center $\displaystyle C\left(\tfrac{5}{4}a,\:0\right)$ and radius $\displaystyle r\,=\,\tfrac{3}{4}a$