Finding distance between 2 points in circle (edge to point)

Question

So I need to get the distance from an edge to another point in the circle. I don't really know how to describe the point, so I've provided a visual. How would I get the length of each of the red lines?

Answer 1

You can find the value of $l$ as a function of angle $\alpha$ easily:

$$DF=DE+EG+GF$$

$$2R\sin\frac{90^\circ+2\alpha}{2}=2l+R\sqrt2$$

$$2R\sin(45^\circ+\alpha)=l+\frac R2 \sqrt2$$

$$R\sqrt2 (\sin\alpha+\cos\alpha)=l+\frac R2 \sqrt2$$

$$l(\alpha)=R\sqrt2 (\sin\alpha+\cos\alpha-\frac12)$$

A bigger challenge is to express $l$ as a function of $x$:

$$x^2+l^2=CD^2=(2R\sin\frac\alpha2)^2=4R^2\sin^2\frac\alpha2=2R^2(1-\cos\alpha)\tag{2}$$

$$\frac{x}{\sin\angle CDE}=\frac{l}{\sin\angle ECD}$$

$$\frac x {\sin(45^\circ+\frac\alpha2)}=\frac l {\sin(45^\circ-\frac\alpha2)}$$

$$\frac xl=\frac{\sin(45^\circ+\frac\alpha2)}{\sin(45^\circ-\frac\alpha2)}$$

$$\frac xl=\frac{\cos\frac\alpha2+\sin\frac\alpha2}{\cos\frac\alpha2-\sin\frac\alpha2}$$

$$\frac xl=\frac{1+\tan\frac\alpha2}{1-\tan\frac\alpha2}\tag{3}$$

Now solve (3) for $\tan\frac\alpha2$cand you get:

$$\tan\frac\alpha2=\frac{x-l}{x+l}\tag{4}$$

Replace (4) into (2):

$$x^2+l^2=2R^2(1-\cos\alpha)=2R^2(1-\frac{1-\tan^2\frac\alpha2}{1+\tan^2\frac\alpha2})=2R^2(1-\frac{1-(\frac{x-l}{x+l})^2}{1+(\frac{x-l}{x+l})^2})\tag{5}$$

After simplification the equation becomes:

$$x^2+l^2=R\sqrt2(x-l)$$

This quadratic equation has only one positive solution for $l$ in terms of $x$:

$$l(x)=\frac{\sqrt 2}{2}(\sqrt{R^2+2\sqrt2 Rx-2x^2}-R)$$

Answer 2

Simplest would be to try and get the equation (red) of line. Then using the equation of circle, you can solve the equation of circle and red line, to get point(s) where red line intersects/touches the circumference.