How does the region $(1-x)^2+y^2 \leq K \left (1 - \sqrt {x^2 + y^2} \right )$ look like?

Question

How does the region $(1-x)^2+y^2 \leq K \left (1 - \sqrt {x^2 + y^2} \right )$ look like?

I drew few demo regions in desmos for different values of $K$ and every time it seems like that the region is bounded by a circle contained in the unit disk except for $K = 1$ which gives rise to a region bounded by a balloon like shape. Does the equation $(1-x)^2+y^2 = K \left (1 - \sqrt {x^2 + y^2} \right )$ represent a circle for $K \neq 1\ $? I am very curious to know that.

Thanks for your time.

Answer 1

We may as well ask about the boundary of the region, namely, the curve $C_K$ defined by $$(1 - x)^2 + y^2) = K \left(1 - \sqrt{x^2 + y^2}\right) .$$ The following plot shows $C_K$ for various values of $K$ along with the limiting circle $x^2 + y^2 = 1$.

As is apparent from some of the curves, no, the boundary is not a circle: Rearranging the equation gives that $C_K$ is some subset of the quartic algebraic curve $$F_K(x, y) := [(x - 1)^2 + y^2 - K]^2 - K^2 (x^2 + y^2) = 0,$$ and for generic values of $K$ it has genus $1$---hence it does not even have a rational parameterization---whereas a circle has genus $0$.

In at least 2 cases $C_K$ has genus $0$ but neither case is a circle:

• $K = 0$: the (real) locus $C_0$ is a single point, $(1, 0)$, and
• $K = 1$: the locus $C_1$ is an arc of the limaçon $r = 1 + 2 \cos \theta$.

One the other hand, for $K \ll 1$ and $K \gg 1$ the curve is approximately a circle, and some algebra suggests why.

Rearranging gives $$1 - \sqrt{x^2 + y^2} = \frac{(x - 1)^2 + y^2}{K} .$$ The right-hand side is positive, so $x^2 + y^2 \leq 1$, hence the left-hand side is $\leq \frac{4}{K}$, so for $K \gg 1$ we have $1 - \sqrt{x^2 + y^2} \approx 0$, and $1 - \sqrt{x^2 + y^2}$ defines the unit circle.

For $K \ll 1$, changing coordinates to $(u, y)$, $x = u + \left(1 - \frac{K}2\right)$, and forming the power series at $u = y = K = 0$ gives that $$0 = u^2 + y^2 - \frac{K^2}{4} + E,$$ where $E$ is an error whose terms have total degree at least $3$ in $u, y, K$, so that $u^2 + y^2 \approx \left(\frac{K}2\right)^2$, and $u^2 + y^2 = \left(\frac{K}2\right)^2$ defines the circle of radius $\frac{K}2$ centered at $(u, y) = (0, 0)$, i.e., at $(x, y) = \left(1 - \frac{K}2, 0\right)$.

Remark The original curve has a compact polar representation:

$$r^2 + (K - 2 \cos \theta) r + (1 - K) = 0$$