Mathematics, 26.07.2019 16:00, fluffykitty21
If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?
Answers: 2
![\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}](/tpl/images/0135/5195/228bf.png)
![\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](/tpl/images/0135/5195/2bad2.png)
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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form e...
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