# Scaling points in circle

I have a set of points that fall inside of a circle with a radius of $1$ and a center of $(0, 0)$. I want to know how to scale those points so that they all have a radius between $0.5$ and $0.75$.

For example, if I have $x = 1$, $y = 0$ then $x_1 = 0.75$, $y_1 = 0$. But that is an easy case on the circumference. What would $x = 0.1$, $y = -0.1$ be and how would you find it?

For any point $(x,y)$, consider the transformation: $$f(x,y)=\left(0.25+\dfrac{0.5}{\sqrt{x^2+y^2}}\right)(x,y)$$
Notice that the distance from this new point to the origin is: $$\left(0.25+\dfrac{0.5}{\sqrt{x^2+y^2}}\right)\sqrt{x^2+y^2}=0.25\sqrt{x^2+y^2}+0.5$$
Hence, since $0 \leq \sqrt{x^2+y^2} \leq 1$, it follows that $0 \leq 0.25\sqrt{x^2+y^2} \leq 0.25$ which implies that $0.5 \leq 0.25\sqrt{x^2+y^2}+0.5 \leq 0.75$. Indeed, observe that this mapping yields:
$$(1,0) \to \left(0.25+\dfrac{0.5}{\sqrt{1^2+0^2}}\right)(1.0)=0.75(1,0)=(0.75,0) \\$$ as well as \begin{align*} (0.1, -0.1) \to \left(0.25+\dfrac{0.5}{\sqrt{0.1^2+(-0.1)^2}}\right)(0.1, -0.1) &=\left(\dfrac14+\dfrac5{\sqrt2} \right)(0.1, -0.1) \\ &\approx(0.37855,-0.37855) \\ \end{align*}
Hint: If we write $x = r \cos \theta$ and $y = r \sin \theta$, we have that $x^2 + y^2 = r^2$. Can you scale $x$ and $y$ by something to make $0.5^2 \leq x^2 + y^2 \leq 0.75^2$?