power of point

Theorem.

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 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 $P{A}^{\prime }$ and $P{B}^{\prime }$ the segments of another secant.  Then the triangles $PA{B}^{\prime }$ and $P{A}^{\prime }B$ are similar since they have equal angles, namely the central angles $\mathrm{\angle }AP{B}^{\prime }$ and $\mathrm{\angle }BP{A}^{\prime }$ and the inscribed angles $\mathrm{\angle }PA{B}^{\prime }$ and $\mathrm{\angle }P{A}^{\prime }B$.  Thus we have the proportion (http://planetmath.org/ProportionEquation)

$$\frac{PA}{P{A}^{\prime }}=\frac{P{B}^{\prime }}{PB},$$

which implies the asserted equation

$$PA\cdot PB=P{A}^{\prime }\cdot P{B}^{\prime }.$$

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

$$K(x,y):={(x-x_0)}^2+{(y-y_0)}^2-r^2=0.$$

Then, by the Pythagorean theorem, we obtain

$p^2={(a-x_0)}^2+{(b-y_0)}^2-r^2$(1)

if $P$ is outside the circle and

$p^2=r^2-({(a-x_0)}^2+{(b-y_0)}^2)$(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

$$K(a,b)\equiv {(a-x_0)}^2+{(b-y_0)}^2-r^2.$$