Question

The locus of a point which divides the join of $A(-1,1)$ and a variable point $P$ on the circle $x^2+y^2=4$ in the ratio $3:2$ meets the line $y=x$ at point:

• A. $(\sqrt{\frac{14}{25}},\sqrt{\frac{14}{25}})$
• B. $(\sqrt{\frac{8}{25}},\sqrt{\frac{8}{25}})$
• C. $(-\sqrt{\frac{14}{25}},-\sqrt{\frac{14}{25}})$
• D. $(-\sqrt{\frac{8}{25}},-\sqrt{\frac{8}{25}})$

Answer 1

Answer: (A), (C)

The correct options are
A $(-\sqrt{\frac{14}{25}},-\sqrt{\frac{14}{25}})$
C $(\sqrt{\frac{14}{25}},\sqrt{\frac{14}{25}})$

Let the coordinates of $P$ be $(2\cos \theta ,2\sin \theta )$ and those of the point which divides $AP$ in the ratio $3:2$ be $(h,k)$.
Then,

$h=\frac{6\cos \theta -2}{5}$ and $k=\frac{6\sin \theta +2}{5}$
Eliminating $\theta$ we get
${\left(\frac{5h+2}{6}\right)}^2+{\left(\frac{5k-2}{6}\right)}^2=1$
or $25(h^2+k^2)+20(h-k)-28=0$

Hence, the locus of $(h,k)$ is $25(x^2+y^2)+20(x-y)-28=0$

which meets the line $y=x$ at $50x^2-28=0$

$\Rightarrow x=y=\pm \sqrt{\frac{14}{25}}$

Hence, the points are $(\sqrt{\frac{14}{25}},\sqrt{\frac{14}{25}})$ and $(-\sqrt{\frac{14}{25}},-\sqrt{\frac{14}{25}})$.