# power of point

Proof.  Let $PA$ and $PB$ be the segments of a secant and $PA^{\prime}$ and $PB^{\prime}$ the segments of another secant.  Then the triangles  $PAB^{\prime}$ and $PA^{\prime}B$ are similar   since they have equal angles, namely the central angles  $\angle APB^{\prime}$ and $\angle BPA^{\prime}$ and the inscribed angles $\angle PAB^{\prime}$ and $\angle PA^{\prime}B$.  Thus we have the proportion (http://planetmath.org/ProportionEquation)

 $\frac{PA}{PA^{\prime}}=\frac{PB^{\prime}}{PB},$

which implies the asserted equation

 $PA\cdot PB=PA^{\prime}\cdot PB^{\prime}.$

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 (http://planetmath.org/SquareOfANumber) 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 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

 $K(x,\,y):=(x-x_{0})^{2}+(y-y_{0})^{2}-r^{2}=0.$
 $\displaystyle p^{2}=(a-x_{0})^{2}+(b-y_{0})^{2}-r^{2}$ (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

 $K(a,\,b)\equiv(a-x_{0})^{2}+(b-y_{0})^{2}-r^{2}.$
Title power of point PowerOfPoint 2013-03-22 15:07:02 2013-03-22 15:07:02 PrimeFan (13766) PrimeFan (13766) 15 PrimeFan (13766) Theorem msc 51M99 power of the point power of a point InversionOfPlane VolumeOfSphericalCapAndSphericalSector