Mathematics, 23.07.2019 10:00 maddigrace3
If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the standard form equation of the circle?
Answers: 1
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -8 &,& 0~) % (c,d) &&(~ -12 &,& 2~) \end{array}\\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-12-(-8)]^2+[2-0]^2}\implies d=\sqrt{(-12+8)^2+(2-0)^2} \\\\\\ d=\sqrt{(-4)^2+2^2}\implies d=\sqrt{20}\qquad \qquad \qquad \qquad \stackrel{\textit{so the radius is then}}{\cfrac{\sqrt{20}}{2}}](/tpl/images/0123/0417/09273.png)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-10}{ h},\stackrel{1}{ k})\qquad \qquad radius=\stackrel{\frac{\sqrt{20}}{2}}{ r}\\\\ -------------------------------\\\\\ [x-(-10)]^2+[y-1]^2=\left( \frac{\sqrt{20}}{2} \right)^2\implies (x+10)^2+(y-1)^2=\cfrac{(\sqrt{20})^2}{2^2} \\\\\\ (x+10)^2+(y-1)^2=\cfrac{20}{4}\implies (x+10)^2+(y-1)^2=5](/tpl/images/0123/0417/3a209.png)
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