Poles, Polars and Orthogonal Circles

Given circles $A(BD)$ and point $C(BD)$, with centers at $A$ and $C$, meeting in points $B$ and $D$, let $E$ be any point on $BD$. Draw tangents $EF$, $EG$ from $E$ to $A(BD).$

Then $FG$ passes through $C$ if and only if the two circles are orthogonal.

Solution

Below is an applet that is supposed to illustrate the assertion:

When you drag points or lines in the applet, the objects leave a trace. To remove the traces press Reset. If you do, drag point $E$ along $BD.$ You will be able to observe that the lines $FG$ all pass through a fixed point. This point is the pole of chord $BD$ in $A(BD).$ You may surmise that that pole coincides with $C$ only if the two circles are orthogonal.

Given circles $A(BD)$ and point $C(BD)$, with centers at $A$ and $C$, meeting in points $B$ and $D$, let $E$ be any point on $BD$. Draw tangents $EF$, $EG$ from $E$ to $A(BD).$

Then $FG$ passes through $C$ if and only if the two circles are orthogonal.

The polar of a point outside a circle with respect to that circle is defined by the two points of tangency of the tangents from the point to the circle.

Point $E$, therefore, is known to be the polar of chord $FG.$ Since $E$ is chosen on $BD,$ the pole of the latter, by La Hire's theorem, lies necessarily on the polar of the former.

Let's $P$ stands for the polar of $BD.$ Then $PF$ is tangent to $A(BD)$ only if $PF\perp AF,$ implying that $P$ is the center of circle $C(BD),$ or that the two circles are orthogonal.