Why is this point set a circle?

Question

consider a circle in Euclidean plane $E$ and any point $A$ in the interior of the circle. Now consider all secants $s_A$ to the circle through the point $A$. The claim is now that the set of midpoints to all the secants $s_A$ consitutes an circle.

I'm wondering how to prove this interesting fact? What is the best approach?

Best regards

Answer 1

CD is a chord and it passes trough A. F is its midpoint. Since $\triangle OCD$ is isosceles $OF\perp CD$. Then if $E$ is the midpoint of $OA$, $FE = EA$ . This will be true for every chord trough $A$.

Answer 2

Without loss of generality, assume

• The outer circle is centered at the origin (by translation)
• The point $A$ is on the $x$-axis (by rotation)

So the circle has equation

$$x^2 + y^2 = r^2 \tag{T1}$$

If the construction creates an inner circle then:

• The leftmost point of the inner circle is halfway between $(-r, 0)$ and $(A, 0)$ : $L = \frac{A - r}{2}$
• The rightmost point of the inner circle is halfway between $(A, 0)$ and $(r, 0)$ : $R = \frac{A + r}{2}$
• The center point of the inner circle is halfway between $L$ and $R$ : $C = \frac{L + R}{2} = \frac{A}{2}$
• The radius of the inner circle is the distance from the center to the edge : $r_2 = R - C = C - L = \frac{r}{2}$

So if an inner circle is constructed, then it has the equation:

$$\left(x - \frac{A}{2}\right)^2 + y^2 = \left(\frac{r}{2}\right)^2 \tag{T2}$$

Consider a line drawn through point $A$ along vector $\begin{bmatrix} m \\ n \end{bmatrix}$ :

$$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} m \\ n \end{bmatrix} t + \begin{bmatrix} A \\ 0 \end{bmatrix} \tag{T3}$$

Finding the intersection of the outer circle and the line, combining (T1) and (T3) , we get:

$$t_1=\frac{\sqrt{(n^2+m^2)r^2 - n^2A^2} - mA}{n^2+m^2}$$

And finding the intersection of the inner circle and the line, combining (T2) and (T3), we get:

$$t_2=\frac{\sqrt{(n^2+m^2)r^2 - n^2A^2} - mA}{2(n^2+m^2)}$$

So $\frac{1}{2}t_1 = t_2$, so the inner circle of (T2) indeed bisects each line from $(A, 0)$ to the outer circle.