Answer
Expanding $\left( {{\bf{X}} - {\bf{a}}} \right)\cdot\left( {{\bf{X}} - {\bf{b}}} \right) = {c^2}$
we obtain
$||{\bf{X}} - \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)|{|^2} = {c^2} + ||\frac{1}{2}\left( {{\bf{a}} - {\bf{b}}} \right)|{|^2}$
This equation defines a sphere with center ${\bf{m}} = \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)$ and radius $R$, where ${R^2} = {c^2} + ||\frac{1}{2}\left( {{\bf{a}} - {\bf{b}}} \right)|{|^2}$.
Work Step by Step
We have ${\bf{X}} = \left( {x,y,z} \right)$.
Let the components of ${\bf{a}}$ and ${\bf{b}}$ be ${\bf{a}} = \left( {{a_1},{a_2},{a_3}} \right)$ and ${\bf{b}} = \left( {{b_1},{b_2},{b_3}} \right)$, respectively.
Write,
$\left( {{\bf{X}} - {\bf{a}}} \right)\cdot\left( {{\bf{X}} - {\bf{b}}} \right) = {c^2}$
${\bf{X}}\cdot{\bf{X}} - {\bf{a}}\cdot{\bf{X}} - {\bf{X}}\cdot{\bf{b}} + {\bf{a}}\cdot{\bf{b}} = {c^2}$
${{\bf{X}}^2} - \left( {{\bf{a}} + {\bf{b}}} \right){\bf{X}} + {\bf{a}}\cdot{\bf{b}} = {c^2}$
$\left( {{\bf{X}} - \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)} \right)\cdot\left( {{\bf{X}} - \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)} \right) - \frac{1}{4}\left( {{\bf{a}} + {\bf{b}}} \right)\cdot\left( {{\bf{a}} + {\bf{b}}} \right) + {\bf{a}}\cdot{\bf{b}} = {c^2}$
$||{\bf{X}} - \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)|{|^2} = {c^2} + \frac{1}{4}\left( {{\bf{a}} + {\bf{b}}} \right)\cdot\left( {{\bf{a}} + {\bf{b}}} \right) - {\bf{a}}\cdot{\bf{b}}$
$||{\bf{X}} - \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)|{|^2} = {c^2} + \frac{1}{4}\left( {{\bf{a}}\cdot{\bf{a}} + 2{\bf{a}}\cdot{\bf{b}} + {\bf{b}}\cdot{\bf{b}}} \right) - {\bf{a}}\cdot{\bf{b}}$
$||{\bf{X}} - \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)|{|^2} = {c^2} + \frac{1}{4}\left( {{\bf{a}}\cdot{\bf{a}} - 2{\bf{a}}\cdot{\bf{b}} + {\bf{b}}\cdot{\bf{b}}} \right)$
$||{\bf{X}} - \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)|{|^2} = {c^2} + ||\frac{1}{2}\left( {{\bf{a}} - {\bf{b}}} \right)|{|^2}$
Since ${\bf{X}} = \left( {x,y,z} \right)$, ${\bf{a}} = \left( {{a_1},{a_2},{a_3}} \right)$ and ${\bf{b}} = \left( {{b_1},{b_2},{b_3}} \right)$, the last equation is the equation of a sphere centered at $\frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)$ and ${R^2} = {c^2} + ||\frac{1}{2}\left( {{\bf{a}} - {\bf{b}}} \right)|{|^2}$.
Hence, $\left( {{\bf{X}} - {\bf{a}}} \right)\cdot\left( {{\bf{X}} - {\bf{b}}} \right) = {c^2}$ defines a sphere with center ${\bf{m}} = \frac{1}{2}\left( {{\bf{a}} + {\bf{b}}} \right)$ and radius $R$, where ${R^2} = {c^2} + ||\frac{1}{2}\left( {{\bf{a}} - {\bf{b}}} \right)|{|^2}$.
