Show that $\lvert x-x_1\rvert=c\cdot\lvert x-x_2\rvert$ describes an $n-1$ sphere for $0<c<1$ and a hyperplane for $c=1$

Question

My attempt:

For $c=1$, it is easy to visualize a 2D plane lying between $x_1$ and $x_2$ and simplifying the equation$\lvert x-x_1\rvert=\lvert x-x_2\rvert$ gives $$x\cdot(x_1-x_2)=\dfrac{(\lvert x_1\rvert^2-\lvert x_2\rvert ^2)}{2}$$

Which is an equation of a hyperplane similar to $z\cdot x=c$ where $z=x_1-x_2$

How do I visualize the $n-1$ sphere when $0 \lt c\lt1$ and proceed with it?

Answer 1

Square everything:$$\|x\|^2 -2 x_1 \cdot x +\|x_1\|^2 = c^2(\|x\|^2-2x_2\cdot x+\|x_2\|^2),$$hence $$\|x\|^2 -2 \frac{(x_1-c^2x_2)}{1-c^2}\cdot x+\frac{\|x_1\|^2-c^2\|x_2\|^2}{1-c^2},$$ which leads to $$\left\|x-\frac{(x_1-c^2x_2)}{1-c^2}\right\|^2 =\left\|\frac{(x_1-c^2x_2)}{1-c^2}\right\|^2-\frac{\|x_1\|^2-c^2\|x_2\|^2}{1-c^2}.$$

Now you need to show that the expression on the right hand side is positive, which should be quite easy to do.