Equation of locus Point P$(x, y)$ moves in such a way that its distance from the point $(3, 5)$ is proportional to its distance from the point $(-2, 4)$. Find the locus of P if the origin is a point on the locus.
Answer:
$$(x-3)^2 + (y-5)^2 =  (x+2)^2 + (y-4)^2$$
or, $$10x+2y-14=0$$
or, $$5x+y-7=0$$
but answer given is $$7x^2+7y^2+128x-36y=0$$

$$(x-3)^2 + (y-5)^2 = k\left[ (x+2)^2 + (y-4)^2 \right]$$ where $k$ is a constant
Now $(0,0)$ lies on the locus. 
Therefore $$9+25=k(4+16) \Rightarrow k=\frac{34}{20} = \frac{17}{10}$$
Using this value of $k$ in the equation, we get
$$(x-3)^2 + (y-5)^2 = \frac{17}{10}\left[ (x+2)^2 + (y-4)^2 \right]$$
$$10\left[(x-3)^2 + (y-5)^2 \right]= 17\left[ (x+2)^2 + (y-4)^2 \right]$$
$$10(x^2-6x+9+y^2-10y+25)=17(x^2+4x+4+y^2-8y+16)$$
$$7x^2+7y^2+128x-36y=0$$
