PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (200)
Orphanage
Unclass'd
Unproven (511)
Corrections (9)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: High Entry average rating: No information on entry rating
[parent] power of point (Theorem)
Theorem 1   If a secant of the circle is drawn through a point ($ P$), then the product of the line segments on the secant between the point and the perimeter of the circle is independent on the direction of the secant. The product is called the power of the point with respect to the circle.

Proof. Let $ PA$ and $ PB$ be the segments of a secant and $ PA'$ and $ PB'$ the segments of another secant. Then the triangles $ PAB'$ and $ PA'B$ are similar since they have equal angles, namely the central angles $ \angle APB'$ and $ \angle BPA'$ and the inscribed angles $ \angle PAB'$ and $ \angle PA'B$. Thus we have the proportion

$\displaystyle \frac{PA}{PA'} = \frac{PB'}{PB},$
which implies the asserted equation
$\displaystyle PA\cdot PB = PA'\cdot PB'.$

\begin{pspicture}(-3,-2)(3,3) \psline{-}(-3,-2)(2.6,3) \psline{-}(-3,-2)(2.8,-1.... ...[r](-1.8,-0.9){$A'$} \rput[a](2,-1.1){$B$} \rput[a](1.9,2){$B'$} \end{pspicture}

Notes. If the point $ P$ is outside a circle, then value of the power of the point with respect to the circle is equal to the square of the limited tangent of the circle from $ P$; this square may be considered as the limit case of the power of point where the both intersection points of the secant with the circle coincide. Another extension of the notion power of point is got when the line through $ P$ does not intersect the circle; we can think that then the intersecting points are imaginary; also now the product of the “imaginary line segments” is the same.

Denote by $ p^2$ the power of the point $ P := (a,\,b)$ with respect to circle

$\displaystyle K(x,\,y) := (x-x_0)^2+(y-y_0)^2-r^2 = 0.$
Then, by the Pythagorean theorem, we obtain
$\displaystyle p^2 = (a-x_0)^2+(b-y_0)^2-r^2$ (1)

if $ P$ is outside the circle and
$\displaystyle p^2 = r^2-((a-x_0)^2+(b-y_0)^2)$ (2)

if $ P$ is inside of the circle. If in the latter case, we change the definition of the power of point to be the negative value $ -p^2$ for a point inside the circle, then in both cases the power of the point $ (a,\,b)$ is equal to
$\displaystyle K(a,\,b) \equiv (a-x_0)^2+(b-y_0)^2-r^2.$



"power of point" is owned by PrimeFan. [ full author list (2) | owner history (2) ]
(view preamble)

View style:

See Also: inversion of plane, volume of spherical cap and spherical sector

Other names:  power of the point, power of a point

This object's parent.

Attachments:
chordal (Result) by PrimeFan
Log in to rate this entry.
(view current ratings)

Cross-references: negative, Pythagorean theorem, imaginary, line, intersection, limit, limited tangent, square, equation, implies, inscribed, central angles, angles, similar, triangles, segments, proof, perimeter, line segments, product, point, circle, secant
There are 3 references to this entry.

This is version 12 of power of point, born on 2005-03-08, modified 2007-09-11.
Object id is 6854, canonical name is PowerOfPoint.
Accessed 3267 times total.

Classification:
AMS MSC: 51M99 (Geometry :: Real and complex geometry :: Miscellaneous)
Pending Errata and Addenda
None.
[ View all 2 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | prove | add result | add corollary | add example | add (any)