Question

The line joining $(5,0)$ to $(10\cos \theta ,10\sin \theta )$ is divided internally in the ratio $2:3$ at $P$, then the locus of $P$ is

• A. $x^2+2xy+y^2-6x=0$
• B. $x+y-3=0$
• C. $(x-3{)}^2+y^2=16$
• D. $x^2=y-3$

Answer 1

The correct option is C $(x-3{)}^2+y^2=16$

Let $(x,y)$ be point $P$

$\therefore x=\frac{2\left(10\cos \theta \right)+3\left(5\right)}{5}$

$\therefore y=\frac{2\left(10\cos \theta \right)+3\left(0\right)}{5}$

$\frac{5y}{10}=\sin \theta ,\frac{y}{4}=\sin \theta$

$\frac{5x-15}{20}=\cos \theta$

$\frac{x-3}{4}=\cos \theta$

${\cos }^2\theta +{\sin }^2\theta =1;\frac{y^2}{16}+\frac{x^2-6x+y}{16}=1$

$(x-3{)}^2+y^2=16$

Low of $P$