$ \text{ The circle can be written as } x^2 + y^2 - 6y + 9 = 0 \text{is } x^2 + (y-3)^2 = 3^2 \\ $

$ \text { Hence the centre is (0,3) and radius 3.} \\ $

$ \text { The Parabola } x^2 = 4y \text{ is symmetrical about the y-axis. The two curves intersect where } \\ $

$ 4y + y ^2 = 6y \Longrightarrow y^2 -2y = 0 \Longrightarrow y(y-2) = 0 \\ $

$ y= 0 \text{ or }y = 2 \\ $

$ \text {when y = 0, x = 0, y = 2, x =} \pm\sqrt 2 \\ $

$ y = \frac{x^2} {4} \Longrightarrow \frac{dy}{dx} = \frac{x}{2} \\ $

$ \text{Required Length} = 2\int_0^{2 \sqrt2} \sqrt { 1 + (\frac{dy}{dx})^2 } dx \\ $

$ = 2\int_0^{2\sqrt 2} \sqrt { 1 + \frac{x^2}{4}} dx \\ $

$ = 2\int_0^{2\sqrt 2} \sqrt {( x^2 + 4)} dx \\ $

$ = \left [ \frac{x}{2} \sqrt {(x^2 + 4) } + \frac{4}{2} log ( x + \sqrt {(x^2 + 4)} ) \right]_0^{2\sqrt2} \\ $

$ = \sqrt 2 . \sqrt (12) + 2 log (2\sqrt2 + \sqrt(12) ) - 2log2 \\ $

$ = 2 \left [ \sqrt 6 + log (\sqrt 2 + \sqrt 3) \right ] \\ $