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

Mathematics

2 answers:

jenyasd209 [6]3 years ago

5 0

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

zhannawk [14.2K]3 years ago

4 0

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

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

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