Why does this would eventually simplify into the original circle equation? I was trying to solve this problem:
The point $A$ has coordinates $(5, 16)$ and the point $B$ has coordinates $(-4,4)$. The variable $P$ has coordinate $(x,y)$ and moves on a path such that $ AP = 2BP$. Show that the Cartesian equation of the path of the $P$ is:
$$(x+7)^2 +y^2 = 100\tag{*}$$
So, what I did is to found a point on the circular path, and shows that the relationship holds. However, the actual answer is to let:
$$(x-5)^2 +(y-16)^2 = 4(x+4)^2 +4(y-4)^2$$
This also just means $AP=2BP$.
However, it will simplify to the original circle equation $(*)$. I literally don't know why. Even if $AP=2BP$ , why would this simplify to $(*)$. What's the mechanism behind this.  
Thank you very much for you guys reply. 

I'll suppose you use the usual euclidian distance between two points. Then the distance between $A$ and $P$ is
$$d(A, P) =\sqrt{(x-5)^2+(y-16)^2}$$
Same goes for the distance between $B$ and $P$.
$$d(B, P)=\sqrt{(x-(-4))^2+(y-4)^2}$$
Since $d(A, P)=D(B, P)$, we have
$$\sqrt{(x-5)^2+(y-16)^2} =2\times\sqrt{(x+4)^2+(y-4)^2}$$
Squaring each side
$$(x-5)^2 +(y-16)^2 = 4\left((x+4)^2 +(y-4)^2\right)$$
Developping each squares
$$x^2-10x+25+y^2-32y+256 = 4x^2+32x+64+4y^2-32y+64$$
Bring all variables to the right, the rest to the left.
$$153=3x^2+42x+3y^2$$
Since the coefficients of $x^2$ and $y^2$ are the same, it is a circle. Divide every term by $3$.
$$51=x^2+14x+y^2$$
Complete the square for $x$
$$51+49=x^2+14x+49+y^2$$
$$100=(x+7)^2+y^2$$
