Circle equation generally:
(x-a)^2 + (y-b)^2 = r^2
The center is in x-axis (b = 0), giving:
(x-a)^2 + y^2 = 1
((sqrt(2)/2)-a)^2 + (sqrt(2)/2)^2 = 1
((sqrt(2)/2)-a)^2 + (1/2) = 1
(sqrt(2)/2)-a = ±sqrt(1/2)
(sqrt(2)/2)-a = sqrt(1/2)
a = 0
So: x^2 + y^2 = 1
(sqrt(2)/2)-a = -sqrt(1/2)
a = sqrt(2)
So: (x-sqrt(2))^2 + y^2 = 1
write the equation of the circle that satisfies each set of conditions.
the center of the circle is on the x-axis, its radius is 1, and it passes through the point (square root of 2 over 2, square root of two over 2)
1 answer
1 answer