## differential geometry

Given $$\{(u,v)\inR^2:u^2+v^2<1\}$$ with metric $$E = G =\frac{4}{(1-u^2-v^2)^2}$$ and $$F = 0$$. How can I show that a Euclidean circle centered at the origin is a hyperbolic circle?

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 Note: rotations about the origin preserve this metric.
 Given $$\{(u,v)\inR^2:u^2+v^2<1\}$$ with metric $$E = G =\frac{4}{(1-u^2-v^2)^2}$$ and $$F = 0$$. With a Euclidean circle centered at the origin with radius r, how can I find the hyperbolic radius by integrating $$\sqrt(E(u')^2+2Fu'v'+G(v')^2)$$, what parametrized curve should I use?