7.7
The Power of the Point
Every point in the plane of a given circle has a special number associated with it called the power of the point with respect to the given circle. This is due to the fact that if a line is drawn through any point $P$ intersecting the circle in two places, the product of the distances measured from $P$ to those two places is constant — it does not change when the line changes. This is true whether $P$ is inside the circle or outside the circle (if $P$ is on the circle, one of the distances is 0 so the product of the two distances is 0). The constant is called the power of the point $P$.
- Theorem 7.19:
The Two-Secant Power Theorem
If secants $\overline{PAB}$ and $\overline{PCD}$ are drawn to a circle from an external point $P$, then $PA\cdot PB=PC\cdot PD$.
Given:
$\overline{PAB}$ and $\overline{PCD}$ are secants to a circle. Prove: $PA\cdot PB=PC\cdot PD$ Proof: Draw chords $\overline{AD}$ and $\overline{BC}$. In $\triangle PDA$ and $\triangle PBC$, there is a common angle, $\angle P$, and there are two inscribed angles, $\angle 1$ and $\angle 2$, that intercept the same arc. These two triangles are similar. Hence $$\dfrac{PD}{PA}=\dfrac{PB}{PC}.$$ Cross mulitplication gives $$PA\cdot PB=PC\cdot PD.$$ |
It is absolutely critical in using this theorem that all distances are measured from the common point $P$. The theorem does not say that the product of the segments on one secant is equal to the product of the segments on a second secant!
Example: Determine the value
of $x$ in the figure to the right. Solution: We have $PA=6$, $PB=12$, $PC=8$, and $PD=8+x$. Then we obtain the following. \begin{align} 6\cdot 12&=8(8+x)\\ 72&=64+8x\\ x&=1 \end{align} |
The situation with a secant and a tangent is exactly the same as the one with two secants. If we think of rotating secant $\overline{PAB}$ clockwise about point $P$ in the figure above until points $C$ and $D$ coincide, then the two distances $PC$ and $PD$ are the same, and we obtain $PA\cdot PB=(PC)^2$. This is an intuitive explanation of what is happening, but a formal similar triangle proof is not hard.
- Theorem 7.20:
The Secant-Tangent Power Theorem
If $\overline{PAB}$ is a secant to a circle and $\overline{PT}$ is a tangent segment to the same circle, then $PA\cdot PB=(PT)^2$.
Example: In the
figure, $\overline{PT}$ is a tangent segment. Determine the value
of $x$. Solution: Now $PA=x$ and $PB=x+5$. We have the following algebraic solution. \begin{align} x(x+5)&=6^2\\ x^2+5x&=36\\ x^2+5x-36&=0\\ (x+9)(x-4)&=0\\ x&=-9\text{ or }4 \end{align} Since $x$ is a length, it must be positive. Therefore $x=4$. |
The power of a point $P$ lying in the interior of a circle is found by taking any chord $\overline{AB}$ through $P$ and forming the product $PA\cdot PB$. Again, as in the last two theorems, all distances are measured from the point $P$.
- Theorem 7.21:
The Two-Chord Power Theorem
If two chords are drawn through the same point in the interior of a circle, then the product of the lengths of the segments on one chord is equal to the product of the lengths of the segments of the other.
Restatement:
In the figure, $PA\cdot PB=PC\cdot PD$. The proof is left as an exercise. |
Example: A chord of length 7 intersects a diameter of length 8 at a point 2 units from the center of the circle. Find the lengths of the segments of the chord.
Solution: The segments on the diameter have lengths 2 and 6. On the other chord, let $x=$ the length of one of the segments; then $7-x=$ the length of the other segment.
\begin{align}
x(7-x)&=2\cdot 6\\
7x-x^2&=12\\
0&=x^2-7x+12\\
0&=(x-3)(x-4)\\
x&=3\text{ or }4
\end{align}