# Inverse with respect to a given circle

Determine the inverse with respect to a given circle $g:\mathbb{R}^{2} \to \mathbb{R}^{+}, g(x,y)=x^{2}+y^{2}$.

I have looked around for non geometric derivations without finding any of value. Anyone willing to lend a helping hand?

• This is not an injective function (not 1-1), so $g^{-1}$ is not a function. For example, $g^{-1}(1)$ contains every pair $(\pm t,\pm\sqrt{1-t^2})$ for $0\leq t\leq 1$ (with all combinations of sign indicated). By $\mathbb R^+$ you must necessarily mean $[0,\infty)$. – MPW Oct 15 '14 at 20:58
• a circle of radius $r$ is the set $(r \cos(\alpha), r \sin(\alpha))$. does this help? – mm-aops Oct 15 '14 at 20:58
• @mm-aops I am looking for a place where I can find a derivation for the result which figures here mathworld.wolfram.com/CircleInverseCurve.html and here math.stackexchange.com/questions/431652/… without the use of a geometric argument. – comPl9121 Oct 15 '14 at 21:07
• so you don't mean 'inverse' in the sense 'inverse of a function' but 'inverse' in the geometric meaning, i.e. inverse with respect to a given circle. you should probably include it in the body of the question cause it seems a bit amigous the way it's written – mm-aops Oct 15 '14 at 21:53
• @mm-aops Thanks for the suggestion, I have changed the question accordingly. – comPl9121 Oct 16 '14 at 0:20