construct a path between $(-1,0)$ and $(0,2)$

Question

So we are given a region $S$ which is above the $x$-axis and between the semicircle of radius $1$ and $2$ centered at the origin. We are asked to construct a path that connect the point $(-1,0)$ and $(0,2)$ and then use this concept to show that $S$ is path connected.

I have found a parabola using the point $(-1,0), (1,0)$ and $(0,2)$ but I'm not sure if this parabola stays in the region $S$, and our professor says that an ellipse is a good choice but too fancy for this question...

Is there any simpler curve or line that can be constructed to connect these two points? and how should I use these points to show that $S$ is path connected ?

Answer 1

From $(-1,0)$, draw a line straigt up, and from $(0,2)$, draw the line to $(-2,0)$. These two lines intersect, and now you should see what path you need.

Answer 2

Your parabola must be $y=2(1-x^2)$. Does it lie inside $S$ ?

As long as $-1\le x\le1$, $y\ge0$, so it stays above the $x$ axis.

Now the squared distance from the origin to a point on the parabola is $d^2=x^2+y^2=x^2+4-8x^2+4x^4=4x^4-7x^2+4$.

Finding the roots of the derivative, $$16x^3-14x=0,$$ we have $x=0\lor x^2=\frac78$ (one maximum and two minima).

Then when $x^2=\frac78$, $d^2=\frac{15}{16}<1$, and the parabola enters the inner circle.

Actually you can guess that because the inner circle has vertical tangents at $x=\pm1$, unlike the parabola.

Answer 3

The following always works for points $A,B$ in your annulus (it's adjusted from Michael's attempt that failed because he used the wrong points.: The perpendicular bisector of $AB$ intersects one of the line segments $OA$, $OB$ in a point $P$. The circle around $P$ through $A$ also passes through $B$. One of the two arcs is completely within the annulus (why?)