Answer
$ (r-a) \cdot (r-b)=0$ represents an equation of a sphere with center: $ [(\frac{a_1+b_1}{2}),(\frac{a_2+b_2}{2}),(\frac{a_3+b_3}{2})]$ and radius: $\frac{|a-b|}{2}$
Work Step by Step
Need to proove $ (r-a) \cdot (r-b)=0$
$ (r-a) \cdot (r-b)=r (r-b)-a (r-b)$
$=r \cdot r-(a+b) \cdot r +a \cdot b$
Now, $r \cdot r-(a+b) \cdot r +a \cdot b= \lt x,y,z \gt \cdot \lt x,y,z \gt- ( \lt a_1,a_2,a_3 \gt +\lt b_1,b_2,b_3 \gt ) \cdot \lt x,y,z \gt+ (\lt a_1,a_2,a_3 \gt ) \cdot (\lt b_1,b_2,b_3 \gt) $
$=x^2+y^2+z^2-(a_1+b_1)x-(a_2+b_2)y-(a_3+b_3)z+a_1b_1+a_2b_2+a_3b_3$
$x^2+y^2+z^2-(a_1+b_1)x-(a_2+b_2)y-(a_3+b_3)z+a_1b_1+a_2b_2+a_3b_3=(x-\frac{a_1+b_1}{2})^2+(y-\frac{a_2+b_2}{2})^2+(z-\frac{a_3+b_3}{2})^2+a_1b_1+a_2b_2+a_3b_3-(\frac{a_1+b_1}{2})^2-(\frac{a_2+b_2}{2})^2-(\frac{a_3+b_3}{2})^2=0$
Thus, $ (r-a) \cdot (r-b)=0$ represents an equation of a sphere with center : $ [(\frac{a_1+b_1}{2}),(\frac{a_2+b_2}{2}),(\frac{a_3+b_3}{2})]$ and radius : $\frac{|a-b|}{2}$
