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July 29, 2014

July 29, 2014

Posted by **Mk** on Monday, February 28, 2011 at 3:16am.

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)

- Advanced Math -
**agrin04**, Monday, February 28, 2011 at 5:52amCircle 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 = ħsqrt(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

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