Answer to Question #147913 in Analytic Geometry for Krystal

Question

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.

Answer 1

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

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

$\begin{cases} (-7-a)^2+(0-b)^2 = R^2\\ (7-a)^2+(0-b)^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 49+14a+a^2+b^2 = R^2\\ 49-14a+a^2+b^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 28a = 0\\ 49-14a+a^2+b^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} a = 0\\ 49-14a+a^2+b^2 = R^2\\ (-3-a)^2+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 49+b^2 = R^2\\ 9+(4-b)^2 = R^2\\ \end{cases}\\ \begin{cases} 49+b^2 = R^2\\ 9+16-8b+b^2 = R^2\\ \end{cases}\\ \begin{cases} 49+b^2 = R^2\\ 25-8b+b^2 = R^2\\ \end{cases}\\ 24+8b = 0\\ b = -3\\ R^2 = 58\\ x^2 +(y+3)^2 = 58$

answer: (0, -3)

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