the solution is hère.

The second equation is $c^2+2cz+z^2=x^2+2ax+a^2+y^2+2by+b^2$. After subtracting the given equalities, $2ax+2by=2cz$ and $ax+by=cz$. Multiplying the two equalities, $(a^2+b^2)(x^2+y^2)=c^2z^2$. But by the Cauchy-Schwartz Inequality, $(a^2+b^2)(x^2+y^2) \ge (ax+by)^2 = c^2z^2$. We have equality here, so we must have $y^2 = \frac{b^2}{a^2} \cdot x^2$, and $z^2=x^2+y^2$. These are all the solutions.

Philip_Leszczynski wrote: We have equality here, so we must have $y^2 = \frac{b^2}{a^2} \cdot x^2$, and $z^2=x^2+y^2$. These are all the solutions. I reached \[\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{x^2+y^2}{a^2+b^2}=\frac{z^2}{c^2},\]and \[\frac{x}{a}=\frac{y}{b}=\frac{z}{c}.\]So, $(x,y,z)\in \{(ma,mb,mc) \mid m \in \mathbb{R}\}$.