Equation of locus

Question

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 $$7x^2+7y^2+128x-36y=0$$

Answer 1

$$(x-3)^2 + (y-5)^2 = \lambda((x+2)^2 + (y-4)^2).$$

We express that the curve passes through the orgin:

$$(-3)^2 + (-5)^2 = \lambda((+2)^2 + (-4)^2),$$

hence $$\lambda=\frac{17}{10}.$$

Answer 2

$$(x-3)^2 + (y-5)^2 = k\left[ (x+2)^2 + (y-4)^2 \right]$$ 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 $$(x-3)^2 + (y-5)^2 = \frac{17}{10}\left[ (x+2)^2 + (y-4)^2 \right]$$ $$10\left[(x-3)^2 + (y-5)^2 \right]= 17\left[ (x+2)^2 + (y-4)^2 \right]$$ $$10(x^2-6x+9+y^2-10y+25)=17(x^2+4x+4+y^2-8y+16)$$ $$7x^2+7y^2+128x-36y=0$$

Answer 3

The same answer than Aniket. You have a cercle centered at the point $(-\frac{64}{7},\frac{18}{7})$ and radius $\frac{2\sqrt{5\cdot13\cdot17}}{7}$