Question

If $A\equiv (9,-1)$ and $B\equiv (-9,5)$, then the locus of moving point $P$ such that $|\overline{AP}|:|\overline{BP}|=1:2$, where $|\overline{AP}|$ denotes the length of $AP$, is

• A. $(x-15{)}^2+(y+3{)}^2=160$
• B. $(x+15{)}^2+(y-3{)}^2=160$
• C. $(x-3{)}^2+(y-15{)}^2=160$
• D. $(x-15{)}^2+(y+3{)}^2=180$

Answer 1

The correct option is A $(x-15{)}^2+(y+3{)}^2=160$

Let $P\equiv (x,y)$. Then
$|\overline{AP}|:|\overline{BP}|=1:2$
$\Rightarrow 2|\overline{AP}|=|\overline{BP}|$
$\Rightarrow 4|\overline{AP}{|}^2=|\overline{BP}{|}^2$
$\Rightarrow 4\left(\right(x-9{)}^2+(y+1{)}^2)=(x+9{)}^2+(y-5{)}^2$
$\Rightarrow 3x^2-90x+3y^2+18y+222=0$
$\Rightarrow x^2-30x+y^2+6y+74=0$
$\Rightarrow (x-15{)}^2+(y+3{)}^2=160$
Hence, the equation of the locus of $P$ satisfying the condition is $(x-15{)}^2+(y+3{)}^2=160$