Centre is at the middle or the midpoint of the diameter.
And the formula of midpoint is
[tex] ( \frac{x_{1} +x_{2}}{2},\frac{y_{1} +y_{2}}{2}) [/tex]
For the given question, x1=-4, y1=2, x2=2, y2=-8
So we will get
[tex] ( \frac{-4+2}{2}, \frac{2-8}{2} ) = (-1, -3) [/tex]
And that's the required centre .
Answer: The required co-ordinates of the center of the circle are (-1, -3).
Step-by-step explanation: Given that AB is the diameter of circle T. The co-ordinates of point A are (-4, 2) and the co-ordinates of point B are (2, -8).
We are to find the co-ordinates of the center of this circle.
We know that
the center of a circle is the mid-point of any of its diameter.
The co-ordinates of the midpoint of a line segment with endpoints (a, b) and (c, d) are given by
[tex]\left(\dfrac{a+c}{2},\dfrac{b+d}{2}\right).[/tex]
Therefore, the co-ordinates of the midpoint of diameter AB are
[tex]\left(\dfrac{-4+2}{2},\dfrac{2-8}{2}\right)\\\\\\=\left(\dfrac{-2}{2},\dfrac{-6}{2}\right)\\\\=(-1,-3).[/tex]
Thus, the required co-ordinates of the center of the circle are (-1, -3).
