Question

A circle has a diameter with endpoints at 3 – 5i and –8 + 2i. What is the center of the circle?

Answer 1

well, the center of the circle is half-way between any two opposite endpoints, let's check then the midpoint for these two.

$\bf \begin{cases} 3-5i\\ \qquad (3,-5)\\ -8+2i\\ \qquad (-8,2) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{2}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)$

$\bf \left( \cfrac{-8+3}{2}~~,~~\cfrac{2-5}{2} \right)\implies \left( -\cfrac{5}{2}~,~-\cfrac{3}{2} \right)\implies \left( -2\frac{1}{2}~,~ -1\frac{1}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -2\frac{1}{2}-1\frac{1}{2}~i~\hfill$

Answer 2

Answer:

The center of the circle is the point $(-2.5, -1,5)$, that can be written as $-2.5-1.5i$.

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 $a+bi$ can be identified with a point in the plane: the point with coordinates $(a,b)$.

So, the number $3-5i$ ‘‘is’’ the point $(3,-5)$, and $-8+2i$ ‘‘is’’ the point $(-8,2)$.

Elementary geometry. The center of a circle is the midpoint of all its diameters.

Now, as the diameter of the circle has endpoints $A=(3,-5)$ and $B=(-8,2)$, we only need to find the midpoint of the segment AB. Form analytic geometry we know that this can  be done by the formulas,

$(x_m,y_m) = \left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right) = \left(\frac{3+(-8)}{2},\frac{-5+2}{2}\right) = (-5/2, -3/2) = (-2.5, -1,5)$