Construct a square with vertices on a given point, line, and circle.

Question

How to construct a square ABCD given point C, circle and a line so that point A lies on the line and point D lies on the circle?

Answer 1

Suppose that the problem is solvable.

Rotate a line around $C$ for $45^{\circ}$ and then dilate it with factor ${1\over \sqrt{2}}$. Intersection point between this new line and a circle is a point $D$ (on picture)...

the square construction Blue(solid) is what is given; the green is the solution.

• $CL$ is the perpendicular to the line;
• the blue dotted line is the original line $45^{\circ}$ rotated;
• $CN$ is ${1\over \sqrt{2}}$-times $CM$;
• $LN$ intersecting the circle gives us point $D$;
• $CD$ is the first side of the square.

How to perform this spiral similarity in practice? If point $X$ is on a line and let $XZCY$ is a (positive oriented) square with diagonal $XC$, then $X$ maps to $Y$. So take two pointy $X_1$ and $X_2$ on a line and then $Y_1Y_2$ is this new line.