Question

Find the coordinates of a point on $y$-axis which are at a distance of $5\sqrt{2}$ from the point $P(3,-2,5)$

Answer 1

Let $A(0,b,0)$ be the point on the $y$-axis at a distance of $5\sqrt{2}$ from point $P(3,-2,5)$.
We have given, $AP=5\sqrt{2}$
$\Rightarrow A{P}^2=50$
$\Rightarrow {(3-0)}^2+{(-2-B)}^2+{(5-0)}^2=50$
$\Rightarrow 9+4+b^2+4b+25=50$
$\Rightarrow b^2+4b-12=0$
$\Rightarrow b^2+6b-2b-12=0$
$\Rightarrow (b+6)(b-2)=0$
$\Rightarrow b=-6,2$
Thus, the coordinates of the required points are $(0,2,0)$ and $(0,-6,0)$