Question

What is an equation for a circle if the endpoints of a diameter are at (3, 4) and (-3, -2)?

Answer 1

since we know the endpoints for the circle, the distance between both will just be the diameter of the circle, and its radius is simply half of that.

let's also bear in mind that the midpoint of those endpoints, is the center of the circle.

`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{-2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√((-3-3)^2+(-2-4)^2)\implies d=√((-6)^2+(-6)^2) \\\\\\ d=√(6^2+6^2)\implies d=√(2(6^2))\implies d=6√(2)\implies \stackrel{\textit{radius = half that}}{r=3√(2)}`

`\bf ~\dotfill\\\\ ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{-2})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-3+3}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \stackrel{\textit{center}}{(0,1)} \\\\[-0.35em] ~\dotfill`

`\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{0}{ h},\stackrel{1}{ k})\qquad \qquad radius=\stackrel{3√(2)}{ r} \\\\\\ (x-0)^2+(y-1)^2=(3√(2))^2\implies \\\\\\ x^2+(y-1)^2=9(2)\implies x^2+(y-1)^2=18`