Question

The distances of point $P(x,y)$ from the points $A(1,-3)$ and $B(-2,2)$ are in the ratio $2:3$.
Show that :
$5x^2+5y^2-34x+70y+58=0$

Answer 1

Distance between P & A is :-

$r_1=\sqrt{{(x-1)}^2+{(y+3)}^2}$

and Distance between P & B is:-

$r_2=\sqrt{{(x+2)}^2+{(y-2)}^2}$

Now,

$\frac{r_1}{r_2}=\frac{2}{3}$

$3r_1=2r_2$

$3\sqrt{{(x-1)}^2+{(y+3)}^2}=2\left(\sqrt{{(x+2)}^2+{(y-2)}^2}\right)$

requiring both side,

$9\left[{\left(x-1\right)}^2+{\left(y+3\right)}^2\right]=4\left[{\left(x+2\right)}^2+{\left(y-2\right)}^2\right]$

$9\left[x^2+1-2x+y^2+9+6y\right]=4\left[x^2+4+4x+y^2+4-4y\right]$

$9x^2-18x+9y^2+54y+90=4x^2+16x+4y^2+6y+32$

$5x^2-34x+5y^2+70y-58=0$