Prove that $|z-10|=3|z-2$| is the equation of a circle with radius $3$ and center $1$. 
Prove that $|z-10|=3|z-2$| is the equation of a circle with radius $3$ and center $1$.

I tried the following solution, with no result:  
Considering $z=x+yi$,  
$$|x+yi-10|=3|x+yi-2|$$
$$|(x-10)+yi|=3|(x-2)+yi|$$
$$\sqrt{(x-10)^2+y^2}=3\sqrt{(x-2)^2+y^2}$$  
power of $2$:  
$$(x-10)^2+y^2=9((x-2)^2+y^2)$$  
after simplification, we will have:  
$$8x^2+8y^2-16x-64=0$$ 
or: $$x^2+y^2-2x-8=0$$
I tried $z=x-yi$, as well. But there is no way I can come up with the intended result, that is, radius of $3$ and center of $1$. Any help would be appreciated.

You are almost at end of your exercise. Just notice that 
$$0=x^2+y^2-2x-8=(x-1)^2+y^2-9=|z-1|^2-3^2\Leftrightarrow |z-1|=3$$
where $z=x+iy$.
