Polar of a circle.

Question

So, I was doing some analytical geometry, circles to be precise. There, the concept of polar was introduced as: "If through a point P (within or without the circle) there be drawn any straight line to meet the circle in Q and R, the locus of the point of intersection of tangents at Q and R is called the polar of P; also P is called the pole of the polar." Now, this locus comes out to be of the form of a straight line which passes through the circle if P lies outside the circumference of the given circle. My doubt is, if the polar is a locus of points of intersection of tangents of a circle, how can it pass through the inside of the circle since that would mean that there are intersection points of two tangents of circle lying inside it, which is downright impossible. Are there "imaginary" points of contact to be considered. P.S.: The book I refer to is The Elements of Coordinate Geometry by S.L. Loney, it hasn't left me in doubts yet.

Answer 1

$P$ inside the circle:

$P$ outside the circle:

Since $\triangle PUO\sim\triangle STO$ (two right triangles sharing the acute angle at $O$) $$ \overline{PO}\,\,\overline{TO}=\overline{UO}\,\,\overline{SO}\tag1 $$ Since $\triangle QUO\sim\triangle SQO$ (two right triangles sharing the acute angle at $O$) $$ \overline{UO}\,\,\overline{SO}=\overline{QO}\,\,\overline{QO}\tag2 $$ Therefore, $$ \overline{PO}\,\,\overline{TO}=\overline{QO}^{\large\,2}\tag3 $$ When $P$ is outside the circle, the intersection of the tangents is outside the circle, but it lies on the line perpendicular to $\overline{PO}$ at the point $T$ satisfying $(3)$.