Given the circumference x^2 - 2x + y^2 = 3, find the position of the center P and the value of the Radius. Then find the intercepts with the y axis and the tangent to the circumference at the positive y intercept.

Find center and radiusFrom completing the quadratic:
x^2 - 2x + 1 - 1 + y^2 = 3(x-1)^2 + y^2 = 4
hence, center P(1, 0) and radius R = sqrt(4) = 2
Find y interceptIntercept with y axis, then x = 0hence1 + y^2 = 4y = +- sqrt(3)and x = 0

Find tangent to positive y interceptintercept Px(0, sqrt(3))Generic line: y = mx + qas x = 0, y = q = sqrt(3)Now we find the value of m. If it is tangent, then intercept between line and circumference must be one value.
We therefore consider the system,
(x-1)^2 + y^2 = 4 eq(1)y = mx + sqrt(3) eq(2)
we substitute eq(2) into eq(1),(x-1)^2 + (mx + sqrt(3))^2 = 4
By rearranging we get,x^2 (1+m^2) + (2 sqrt(3) m - 2)x = 0We already considered the solution x = 0,hence we have
x = (2-2sqrt(3)m)/(1+m^2)
As we want the two solutions to be "degenerate", i.e. to coincide as a tangent intercept a circumference in one point only, then we set m = 1/sqrt(3) so that x is 0 again.
End.