Question

From the point $(1,-2,3)$, lines are drawn to meet the sphere $x^2+y^2+z^2=4$ and they are divided internally in the ratio $2:3$. The locus of the point of division is

• A. $5x^2+5y^2+5z^2-6x+12y+2z=0$
• B. $5x^2+5y^2+5z^2=0$
• C. $5x^2+5y^2+5z^2-2xy-3yz-zx-6x+12y+5z+22=0$
• D. None of these

Answer 1

The correct option is D None of these

Suppose any line through the given point $(1,-2,3)$ meets the sphere $x^2+y^2+z^2=4$ in the point $(x_1,y_1,z_1).$
Then, $x_1^2+y_1^2+z_1^2=4$ ...$\left(1\right)$

Now, let the coordinates if the point which divides the join of $(1,-2,3)$ and $(x_1,y_1,z_1)$ in the ratio $2:3$ be $(x_2,y_2,z_2)$. Then we have
$x_2=\frac{2.x_1+3.1}{2+3}$ or $x_1=\frac{5x-3}{2}$
$y_2=\frac{2.y_1+3(-2)}{2+3}$ or $y_1=\frac{5y_2+6}{2}$
$z_2=\frac{2.z_1+3.3}{2+2}$ or $z_1=\frac{5z_2-9}{2}$
Putting the values of $x_1+y_1+z_1$ in $\left(1\right)$, we have
${(5x_2-3)}^2+{(5y_2+6)}^2+{(5z_2-9)}^2=4\times 4$
$\Rightarrow 25(x_2^2+y_2^2+z_2^2)-30x_2+60y_2-90z_2+110=0$
$\therefore$ The locus of $(x_2,y_2,z_2)$
$5(x^2+y^2+z^2)-6(x-2y+3z)+22=0$