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(1)/(2)~,~ -1(1)/(2)\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -2(1)/(2)-1(1)/(2)~i~\hfill`

Answer 2

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

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((x_A+x_B)/(2),(y_A+y_B)/(2)\right) = \left((3+(-8))/(2),(-5+2)/(2)\right) = (-5/2, -3/2) = (-2.5, -1,5)`