If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?
2 answers:
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}](https://tex.z-dn.net/?f=%5Cbf%20%0A~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-8%20%26%2C%26%20-6~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%20-4%20%26%2C%26%20-14~%29%0A%5Cend%7Barray%7D%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-4-%28-8%29%5D%5E2%2B%5B-14-%28-6%29%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-4%2B8%29%5E2%2B%28-14%2B6%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B4%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B80%7D%5Cqquad%20%5Ctextit%7Btherefore%20the%20radius%20is%20%7D%5Cqquad%20r%3D%5Ccfrac%7B%5Csqrt%7B80%7D%7D%7B2%7D)
now, since the diameter goes through the center of the circle, the midpoint of it, will be the center of the circle then,
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](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-6%7D%7B%20h%7D%2C%5Cstackrel%7B-10%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%5Cfrac%7B%5Csqrt%7B80%7D%7D%7B2%7D%7D%7B%20r%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-6%29%5D%5E2%2B%5By-%28-10%29%5D%5E2%3D%5Cleft%28%20%5Cfrac%7B%5Csqrt%7B80%7D%7D%7B2%7D%20%5Cright%29%5E2%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B6%29%5E2%2B%28y%2B10%29%5E2%3D%5Ccfrac%7B%28%5Csqrt%7B80%7D%29%5E2%7D%7B2%5E2%7D%5Cimplies%20%28x%2B6%29%5E2%2B%28y%2B10%29%5E2%3D%5Ccfrac%7B80%7D%7B4%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B6%29%5E2%2B%28y%2B10%29%5E2%3D20)
now, since the diameter goes through the center of the circle, the midpoint of it, will be the center of the circle then,
and since we know the coordinates for the center and the radius, let's plug those two in
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