Write the equation of a circle that passes through point 4 on the Ox axis and through point 8 on the Oy axis, if the center is known to be on the Ox axis.
The general equation of the circle:
(x-x0) ^ 2 + (y-y0) ^ 2 = r ^ 2, where (x0, y0) is the center of the circle, r is the radius of the circle. The circle must pass through points (4.0) and (0.8). This means that when these points are substituted into the equation of the circle, it turns into a true equality. The center of the circle lies on the Ox axis, which means the coordinates of the center (x0, 0)
So, we get the system of equations
(4-x0) ^ 2 = r ^ 2
x0 ^ 2 + 8 ^ 2 = r ^ 2
16-2 * 4 * x0 + x0 ^ 2 = x0 ^ 2 + 64
8×0 = 16-64
8×0 = -48
x0 = -6
Center coordinates (-6.0)
Substitute x0 into the first equation:
(4 – (- 6)) ^ 2 = 10 ^ 2 = r ^ 2, so r = 10
So, we get the equation of the circle
(x + 6) ^ 2 + y ^ 2 = 100
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