Construction of secant with ruler and compass

Question

A is a point outside the cicle $\Omega$. Construct with ruler and compass the secant AB, such as AC=CB with $C,B \in \Omega$ I think C must be the center of a circle with radius AC,or maby AB is the diagonal of a parallelogram with sides AO and r ( O , r center and radius of $\Omega $) and C is the interception of its diagonals. Please help

Answer 1

HINT- First at all, an analytical look at the problem in order to know what we can do. In the figure below it is clear that the angle $\angle BAO$ is decreasing from $\alpha_0$ (when $B$ is a point of tangency) till $0$ (when $B$ is in the line $AO$) which corresponds to the chord $CB$ of the circle $\Omega$ increasing from $0$ till $2r$(the length of the largest chord of $\Omega$). All this obviously has a symmetric below the line $AO$.

Notice that $$AC=CB\iff AC\cos\alpha=CB\cos \alpha$$ It follows that

► if $d(A,O)\gt 3r$ then there is no solution the maximun of $BC$ being $2r$ and $AC$ being always greater than $2r$.

► if $d(A,O)=3r$ then the solution is given by the point $C$ in the line $AO$ and $AC=CB=2r$. Besides the solution coincides with its symmetric.

► if $d(A,O)\lt 3r$ then, because the values of $CB$ change continuously, , there is a solution (to be calculated) by the intermediate value theorem. Calculation gives (median $OC$ in function of the three sides of the triangle, see figure below) $$r^2=\frac 14\left(2r^2+a^2-(2x)^2\right)\iff 4x^2=2r^2-a^2\Rightarrow x=\frac{\sqrt{2r^2-a^2}}{2}$$ Construction with ruler and compass.

Since we know that $ x=\frac{\sqrt{2r^2-a^2}}{2}$ all we need now is to know how to construct the square and the square root of a segment; the difference of two (collinear) segments and the division by $ 2 $ is immediate. We need for this to construct the perpendicular to a line in a point and the circumcircle of a triangle. I assume you know these quite known constructions.

When you have constructed the segment $x$ you can finish with radius $x$ and center $A$. The intersection with the circle $\Omega$ determines the point solution $C$ for which $AC=CB$ where clearly the point $B$ is obtained by extending the line AC.

Answer 2

Let $AT$ be a tangent to the circle. Recall that $AT*AT = AB*AC = 2 AC*AC$. Therefore $AC = \frac{AT}{\sqrt{2}}$, a side of a square with a diagonal equal to $AT$.