# Circle and a point inside it

This is a question that follows from article $$4160$$ of George Carr's book. He states that:

If $$(x, y)$$ be a point P in a circle $$S$$(where $$S={(x-a)^2}+{(y-b)^2}-r^2$$), then $$S$$ becomes minus the ordinate drawn through $$P$$ at right angles to the radius through $$P$$.

My doubt is :

By ordinate, does he mean the $$y$$-intercept of the line drawn perpendicular to the radius through $$P$$?

• Please cite the original, since the circle becomes minus the ordinate makes no sense Commented Apr 1, 2023 at 16:39

I think the word ordinate is used in an old sense related to conic sections. For example the Wordnik online dictionary gives one possible definition of 'ordinate'

Any one of a set of parallel chords of a conic in relation to the diameter bisecting them. What in this sense was called semiordinate is now usually called ordinate.

In your document I think it simply means that if $$P$$ is inside the circle, the half chord through $$P$$ perpendicular to the radius has length squared equal to $$-S$$.

Note that $$S$$ is not the circle, it is the quantity $$(x-a)^2+ (y-b)^2- r^2$$. The result cited here is a trivial consequence of the Pythagorean theorem.

This book "Synopsis of Elementary Results in Pure Mathematics" by George S. Carr has been published in 1886 and uses some old vocabulary (see "More explanations" below).

Let us rephrase the issue in a different way.

Let $$(C)$$ be the circle with center $$(a,b)$$ and radius $$r$$, i.e., with equation :

$$(x-a)^2+(y-b)^2-r^2=0$$

This circle induces two regions in the plane:

• the set of points $$(x,y)$$ such that $$(x-a)^2+(y-b)^2-r^2<0$$ which is the interior of (C).

• the set of points $$(x,y)$$ such that $$(x-a)^2+(y-b)^2-r^2>0$$ which is the exterior of (C).

This article "4160" expresses the equality between two ways of describing the power of an interior point $$P$$ with respect to circle $$(C)$$:

$$\mathfrak{P}(P,(C))=\begin{cases}(x-a)^2+(y-b)^2-r^2<0\\ \underbrace{\vec{PM}.\vec{PM'}}_{\text{dot product}}=\underbrace{-PM.PM'}_{\text{opposite of lengths' product}}=-PM^2\end{cases}$$

(see the answer by Blue to this question).

More explanations : The way one must understand the word "ordinates" is probably due to the fact that $$OP$$ is implicitly taken as a natural abscissa axis whereas the other one $$MM'$$ called traditionaly "conjugate", in this case the orthogonal axis to $$OP$$. Said otherwise : consider all lines parallel to $$MM'$$ : their midpoints are aligned along a line which is line $$OP$$ like in this recent question here.