Question

The diameter of a circle has endpoints (-10,5) and (-4,11)

Part A- find the center of the circle
Part B- find the distance of the radius
Part C- write the equation of the circle

Answer 1

well, since we know the endpoints for the diameter, its midpoint will be where the center of the circle is located, so

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{11}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -4 -10}{2}~~~ ,~~~ \cfrac{ 11 + 5}{2} \right)\implies \left(\cfrac{-14}{2}~~,~~\cfrac{16}{2} \right)\implies \stackrel{center}{(-7~~,~~8)}`

and if we get the distance between those endpoints, and take half of that, that'd be the radius of it.

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{11})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{diameter}{d}=√([-4 - (-10)]^2 + [11 - 5]^2)\implies d=√((-4+10)^2+6^2) \\\\\\ d=√(6^2+6^2)\implies d=√(72)~\hfill \stackrel{radius=half~that}{\cfrac{√(72)}{2}} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-7}{ h},\stackrel{8}{ k})\qquad \qquad radius=\stackrel{(√(72))/(2)}{ r} \\\\\\\ [x-(-7)]^2~~ + ~~[y-8]^2~~ = ~~\left( (√(72))/(2) \right)^2\implies (x+7)^2~~ + ~~(y-8)^2~~ = ~~18`