Question

Given A (-5, 2) and B is the point on the locus whose equation is $x^2+y^2-2x+4y+8=0$. If the point P divides segment AB externally in the ratio 2:1, find the equation of locus of P.

Answer 1

Given $(-5,2)\to A$

$B(x,y)$ where $x^2+y^2-2x+4y+8=0$

$\Rightarrow (x^2-2x+1)+(y^2+4y+4)+3=0$

$\Rightarrow (x-1{)}^2+(y+2{)}^2+3=0$

Let P be $(\alpha ,\beta )$

(Refer image)

$m=2n=1$

$\alpha =\frac{2x-1(-5)}{(2-1)},\beta \left(\frac{2y-\frac{1}{2}}{(2-1)}\right)$

$\alpha =2x+5$ ; $\beta =2y-2$

$\Rightarrow x=\left(\frac{\alpha -5}{2}\right)$ and $y=\left(\frac{\beta +2}{2}\right)$

but $(x-1{)}^2+(y+2{)}^2+3=0$

${(\frac{\alpha -5}{2}-1)}^2+(\frac{\beta +2}{2}+2)+3=0$

$(\frac{\alpha -5}{2}-1{)}^2+(\frac{\beta +6}{2}+2)+3=0$

$\therefore$ lower of $P(\alpha ,\beta )$ is $(\alpha -1{)}^2+(\beta +6{)}^2+12=0$