Question #147913

An archeologist found the remains of an ancient wheel, which she then placed on a grid. If an arc of the wheel passes through A(-7, 0), B(-3, 4) and C(7, 0) locate the center of the wheel, and the standard equation of the circle defining its boundary.


1
Expert's answer
2020-12-03T08:00:21-0500

the circle has an equation (xa)2+(yb)2=R2(x-a)^2+(y-b)^2 = R^2 ,

(a, b) - center circle, point A, B, C belong circle

{(7a)2+(0b)2=R2(7a)2+(0b)2=R2(3a)2+(4b)2=R2{49+14a+a2+b2=R24914a+a2+b2=R2(3a)2+(4b)2=R2{28a=04914a+a2+b2=R2(3a)2+(4b)2=R2{a=04914a+a2+b2=R2(3a)2+(4b)2=R2{49+b2=R29+(4b)2=R2{49+b2=R29+168b+b2=R2{49+b2=R2258b+b2=R224+8b=0b=3R2=58x2+(y+3)2=58\begin{cases} (-7-a)^2+(0-b)^2 = R^2\\ (7-a)^2+(0-b)^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 49+14a+a^2+b^2 = R^2\\ 49-14a+a^2+b^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 28a = 0\\ 49-14a+a^2+b^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} a = 0\\ 49-14a+a^2+b^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 49+b^2 = R^2\\ 9+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 49+b^2 = R^2\\ 9+16-8b+b^2 = R^2\\ \end{cases}\\ \begin{cases} 49+b^2 = R^2\\ 25-8b+b^2 = R^2\\ \end{cases}\\ 24+8b = 0\\ b = -3\\ R^2 = 58\\ x^2 +(y+3)^2 = 58

answer: (0, -3)

x2+(y+3)2=58x^2 +(y+3)^2 = 58


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