well, the center of the circle is half-way between any two opposite endpoints, let's check then the midpoint for these two.
[tex]\bf \begin{cases} 3-5i\\ \qquad (3,-5)\\ -8+2i\\ \qquad (-8,2) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{2}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)[/tex]
[tex]\bf \left( \cfrac{-8+3}{2}~~,~~\cfrac{2-5}{2} \right)\implies \left( -\cfrac{5}{2}~,~-\cfrac{3}{2} \right)\implies \left( -2\frac{1}{2}~,~ -1\frac{1}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -2\frac{1}{2}-1\frac{1}{2}~i~\hfill[/tex]
Answer:
The center of the circle is the point [tex](-2.5, -1,5) [/tex], that can be written as [tex] -2.5-1.5i[/tex].
Step-by-step explanation:
Here we will need some facts from elementary geometry, and other from the geometric representation of complex numbers.
Geometrical representation of complex numbers. Recall that a complex number [tex]a+bi[/tex] can be identified with a point in the plane: the point with coordinates [tex](a,b)[/tex].
So, the number [tex]3-5i[/tex] ‘‘is’’ the point [tex](3,-5)[/tex], and [tex]-8+2i[/tex] ‘‘is’’ the point [tex](-8,2)[/tex].
Elementary geometry. The center of a circle is the midpoint of all its diameters.
Now, as the diameter of the circle has endpoints [tex]A=(3,-5)[/tex] and [tex]B=(-8,2)[/tex], we only need to find the midpoint of the segment AB. Form analytic geometry we know that this can be done by the formulas,
[tex](x_m,y_m) = \left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right) = \left(\frac{3+(-8)}{2},\frac{-5+2}{2}\right) = (-5/2, -3/2) = (-2.5, -1,5) [/tex]
