Circle in a Semicircle

Let $A$ be the center of the circle that forms the semicircle, $B$ be any point along the resulting orange curve, $C$ be the point of tangency of the inscribed circle with center $B$ at the flat part of the semicircle, $D$ be the point of tangency of the inscribed circle with center $B$ at the curved part of the semicircle, and $E$ be the intersection of $BC$ and the tangent line through $D$ , as shown below.

Since $AD$ is the radius of the semicircle, and $DE$ is a tangent line of the semicircle, $AD \perp DE$ at $D$ ; and since $BD$ is the radius of the inscribed circle with center $B$ , and $DE$ is the tangent line of the inscribed circle, $BD \perp DE$ at $D$ . Since there is only one perpendicular line that passes through a given point on a line, $A$ , $B$ , and $D$ must lie on the same line.

Then $BC \cong BD$ since they are both radii of the inscribed circle with center $B$ , $\angle ABC \cong \angle EBD$ since they are vertical angles, and $\angle ACB \cong \angle EDB$ since they are both right angles formed by radii and tangent lines. Therefore, $\triangle ABC \cong \triangle EBD$ by angle-side-angle congruence.

Since $\triangle ABC \cong \triangle EBD$ , $AB \cong EB$ , and since $BC \cong BD$ and $AD = AB + BD$ and $EC = EB + BC$ , $AD \cong EC$ . Since $AD$ is the radius of the semicircle, it is constant for any point $B$ chosen along the orange curve, and so $EC$ is also constant. This means that the locus of points of $E$ for any chosen point $B$ along the orange curve is the line $m$ , a line that is parallel to the flat part of the semicircle and perpendicular to $BE$ .

Since $AB \cong EB$ , any point $B$ along the orange curve is equidistant from a focus at point $A$ and a directrix at line $m$ , which is by definition a parabola .

Lets call the semicircle $S$ and the center of the inscribed circle ${P(x, y)}$ .

You can represent the semicircle on a Cartesian coordinate like this (Radius= $R$ ):

$S:y=\sqrt{R^2-x^2}$

Since $R$ is equal to $\overline{OP}$ +(radius of inscribed circle),

and radius of inscribed circle is always equal to y coordinate of $P$ ,

next equation is valid.

$\sqrt{x^2+y^2}+y=R$ $(-R\leq x\leq R)$

Rearrange it a bit, and you get this.:

$\sqrt{x^2+y^2}=R-y$

$x^2+y^2=(R-y)^2$

$x^2+\not{y^2}=\not{y^2}-2Ry+R^2$

$2Ry=-x^2+R^2$

$y=-\frac{1}{2R}x^2+\frac{R}{2}$

which is a general equation of parabola.

The general expression for a conic section in polar is

$r(\theta) = \frac{a}{1+\epsilon \cos(\theta)}$

Where $\epsilon$ is the eccentricity. The type of conic section is determined exclusively by $\epsilon$ .

Let $O$ be the midpoint of the diameter of the semicircle. Then for any inscribed circle $C$ , there is some radius of the semicircle which passes through the inscribed circle's center. Let $a$ be the radius of the semicircle and $r$ be the distance from $O$ to the center of $C$ . Let $\theta$ be the angle formed by a radius of the semicircle and the perpendicular bisector of the diameter of the semicircle. Then

$r+r \cos (\theta) = a \Rightarrow r = \frac{a}{1+ \cos(\theta)}$

This describes a conic section of eccentricity $1$ , which is a $\boxed{\text{parabola}}$ .

For the big semicircle let's call O its center (used as origin) and R its ray. For a chosen inscribed circle let's call C its center and and r its ray

Let's call L the horizontal line $y=R$

$distance(L,C)=R-r=distance(O,C)$

so it's a parabola with focus in O and directrix L