Circle

Slope of the straight line = sqrt (3)

So equation of the line is:

(y - 0)/(x + sqrt 3) = sqrt 3

y = (sqrt 3)x + 3 ... (*)

For the circle, it passes the origin and has its centre on the y-axis, so its equation should be of the form:

x2 + y2 + ky = 0 where k is a constant.

Sub (*) into this equation:

x2 + [(sqrt 3)x + 3]2 + k[(sqrt 3)x + 3] = 0

x2 + 3x2 + 6(sqrt 3)x + 9 + k (sqrt 3) x + 3k = 0

4x2 + (k + 6)(sqrt 3)x + (9 + 3k) = 0 ... (**)

Since the line touches the circle, discriminant of (**) should be zero, i.e.

[(k + 6)(sqrt 3)]2 - 16(9 + 3k) = 0

3(k + 6)2 - 144 - 48k = 0

3k2 + 36k + 108 - 144 - 48k = 0

3k2 - 12k - 36 = 0

k2 - 4k - 12 = 0

(k - 6)(k + 2) = 0

k = 6 or -2

Since the centre of the circle has a positive y-coord, k should be negative and hence k = -2

Thus equation of circle is x2 + y2 - 2y = 0

Let (0 , Y) be the centre :

Then Y / √3 = tan (60/2)°

Y / √3 = √3 / 3

Y = 1

The centre is (0 , 1).

The radius = Y = 1

The required equation is (x - 0)² + (y - 1)² = 1²

==> x² + (y - 1)² = 1