Question

$A(-9,0),B=(-1,0)$ are two points. lf $P$ $(x,y)$ is a point such that $3PB=PA$, then locus of $P$ is

• A. $x^2-y^2=9$
• B. $x^2-y^2=-9$
• C. $x^2+y^2=9$
• D. $x^2+y^2=3$

Answer 1

The correct option is C $x^2+y^2=9$

Let $P(x,y)$ be a point lying on the curve.
Then $3PB=PA$

$3\sqrt{(x+1{)}^2+y^2}=\sqrt{(x+9{)}^2+y^2}$
Squaring both sides give us:
$9(x+1{)}^2+9y^2=(x+9{)}^2+y^2$
$\Rightarrow (3x+3{)}^2-(x+9{)}^2+8y^2=0$
$\Rightarrow (4x+12)(2x-6)+8y^2=0$
$\Rightarrow 8(x+3)(x-3)+8y^2=0$
$\Rightarrow (x+3)(x-3)+y^2=0$
$\Rightarrow x^2-9+y^2=0$
$\Rightarrow x^2+y^2=9$