Question

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?

Answer 1

(x+6)^2 +(y+10)^2 = 20

Answer 2

So, since we know the coordinates for the diameter segment, let's see how long it is, recall that the radius is half of the diameter, thus

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -8 &,& -6~) % (c,d) &&(~ -4 &,& -14~) \end{array}\\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-4-(-8)]^2+[-14-(-6)]^2} \\\\\\ d=\sqrt{(-4+8)^2+(-14+6)^2}\implies d=\sqrt{4^2+(-8)^2} \\\\\\ d=\sqrt{80}\qquad \textit{therefore the radius is }\qquad r=\cfrac{\sqrt{80}}{2}$

now, since the diameter goes through the center of the circle, the midpoint of it, will be the center of the circle then,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -8 &,& -6~) % (c,d) &&(~ -4 &,& -14~) \end{array}~~ % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-4-8}{2}~~,~~\cfrac{-14-6}{2} \right)\implies (-6~,~-10)$

and since we know the coordinates for the center and the radius, let's plug those two in

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-6}{ h},\stackrel{-10}{ k})\qquad \qquad radius=\stackrel{\frac{\sqrt{80}}{2}}{ r} \\\\\\\ [x-(-6)]^2+[y-(-10)]^2=\left( \frac{\sqrt{80}}{2} \right)^2 \\\\\\ (x+6)^2+(y+10)^2=\cfrac{(\sqrt{80})^2}{2^2}\implies (x+6)^2+(y+10)^2=\cfrac{80}{4} \\\\\\ (x+6)^2+(y+10)^2=20$