Question

Ends of a diameter: (-11, -9) and (1, -3) whats the radius and center

Answer 1

well, we know the endpoints of its diameter, so hmmm its center will be the midpoint of those endpoints.

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-11}~,~\stackrel{y_1}{-9})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -11}{2}~~~ ,~~~ \cfrac{ -3 -9}{2} \right)\implies \left( \cfrac{-10}{2}~~,~~\cfrac{-12}{2} \right)\implies \stackrel{center}{(-5~~,~~-6)}`

well, to get its radius, we can simply get the distance between both points and keeping in mind that the radius is half the diameter, we'll take half of that distance.

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-11}~,~\stackrel{y_1}{-9})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√([1 - (-11)]^2 + [-3 - (-9)]^2)\implies d=√((1+11)^2+(-3+9)^2) \\\\\\ d=√(12^2 + 6^2)\implies d=√(180)\implies d=6√(5)~\hfill \underset{half~that}{\stackrel{radius}{3√(5)}}`