Complex Number Question Help

If x+ iy = 1 / (2+ cosB + i sinB ), where x, y, B are real; then show that, when B varies, the point (x, y) moves on a circle whose center is the point (2/3,0).

Invert both sides to get 1/(x + iy) = 2 + cos(B) + i*sin(B) ---->

(1/(x + iy))*((x - iy) / (x - iy)) = 2 + cos(B) + i*sin(B) ----->

(1/(x^2 + y^2))*(x - iy) = 2 + cos(B) + i*sin(B).

Comparing real and complex components gives us that

(1/(x^2 + y^2))*x = 2 + cos(B) and (1/(x^2 + y^2))*(-y) = sin(B).

So cos(B) = (x/(x^2 + y^2)) - 2 and sin(B) = (-y/(x^2 + y^2)).

Now cos^2(B) + sin^2(B) = 1 ----->

(x^2 / (x^2 + y^2)^2) - (4x/(x^2 + y^2)) + 4 + (y^2/(x^2 + y^2)^2) = 1 ---->

x^2 - 4x*(x^2 + y^2) + 4*(x^2 + y^2)^2 + y^2 = (x^2 + y^2)^2 ------>

(x^2 + y^2)*(4x - 1) = 3*(x^2 + y^2)^2, so either x^2 + y^2 = 0 or

4x - 1 = 3*(x^2 + y^2) ----->

x^2 - (4/3)*x + y^2 = 1 ----->

(x^2 - (4/3)*x + (2/3)^2) + y^2 = 1 + (2/3)^2 ----->

(x - (2/3))^2 + y^2 = 13/9,

which is a circle of radius sqrt(13)/3 and center (2/3, 0).

a little correction the final result i got by solving is (x-(2/3))^2 + y^2 = (1/3)^2
hence radius is 1/3.

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