**power of a point**with respect to a

*circle*.

Suppose on a plane we have a point $P$ and a circle $(O)$. Draw a line through $P$ which intersects with the circle at two points $U$ and $V$. Then the value of $$PU \times PV$$ is

*independent of the choice of the line*$PUV$.

This means that if we draw another line through $P$ which cuts the circle at two other points $A$ and $B$ then $$PA \times PB = PU \times PV.$$

This constant value is called

*the power of the point*$P$

*with respect to the circle*$(O)$.

To prove that $PU \times PV$ is

*independent*of the position of the line, we will use similar triangles. We will consider two separate cases: when the point $P$ is

*outside*the circle and when the point $P$ is

*inside*the circle.

When $P$ is outside the circle $(O)$

We consider two triangles $PUB$ and $PAV$. These two triangles are similar triangles because they have two pairs of equal angles. Thus, $$\frac{PU}{PA} = \frac{PB}{PV}.$$

From here we obtain the required equality $$PA \times PB = PU \times PV.$$

When $P$ is inside the circle $(O)$

When $P$ is inside the circle, we also consider the two triangles $PUB$ and $PAV$. These two triangles are similar triangles because they have equal angles. So $$\frac{PU}{PA} = \frac{PB}{PV}.$$

This implies the constancy of the power of the point $P$: $$PA \times PB = PU \times PV.$$

Expressing power of the point $P$ in terms of the distance $PO$ and the circle radius $r$

We will now show how to derive a formula for the power of the point $P$ in terms of the

**distance**$PO$ and the

**radius**$r$ of the circle $(O)$. We again consider two separate cases: when the point $P$ is outside the circle and when the point $P$ is inside the circle.

We will prove that:

- If $P$ is
*outside*the circle then $$PU \times PV = PO^2 - r^2.$$ - If $P$ is
*inside*the circle then $$PU \times PV = r^2 - PO^2.$$

Indeed, if we take $U$ and $V$ as two intersection points of the line $PO$ with the circle $(O)$ then the power of the point $P$ can be calculated as follows:

- When $P$ is
*outside*the circle $$PU \times PV = (PO - r)(PO + r) = PO^2 - r^2.$$ - When $P$ is
*inside*the circle $$PU \times PV = (r - PO)(r + PO) = r^2 - PO^2.$$

So we have shown that the power of the point $P$ can be expressed in terms of the

*distance*$PO$ and the

*radius*$r$ as $$PU \times PV = \pm (PO^2 - r^2).$$

Whether the sign is a

*plus*or a

*minus*in the above formula depends on the position of the point $P$ with respect to the circle. It is a

*plus*if $P$ is

*outside*the circle and it is a

*minus*if $P$ is

*inside*the circle. (

*For the special case*when the point $P$ lies

*on the circle*then the power of $P$ is equal to

*zero*trivially.)

The "correct" definition of the power of a point

Actually our above definition of

**the power of a point**is not really correct. The above definition does not concern about the

**sign**of $PU \times PV$ making it always a positive number.

The "

*correct*" definition of the power of the point $P$ is $$\vec{PU} \times \vec{PV} = PO^2 - r^2.$$

In this definition, the value of the power of the point $P$ can be

*positive*or

*negative*, it depends on the two vectors $\vec{PU}$ and $\vec{PV}$ point to the

*same direction*or

*opposite directions*.

When $P$ is outside the circle, the power of $P$ is a positive number because $\vec{PU}$ and $\vec{PV}$ have the same direction.When $P$ is inside the circle, the power of $P$ is a negative number because $\vec{PU}$ and $\vec{PV}$ point to opposite directions. |

*The power of a point*is a useful bit of information that helps us to know the relative position of the point with respect to the circle. If this value is a

*positive*number then we know that the point is

*outside*of the circle. If this value is

*zero*then the point must lie

*on the circle*. And if this value is

*negative*then we know that the point is

*inside*the circle.

Formula of the power of a point in Cartesian coordinate

In the Cartesian coordinate system $0xy$, each point $P$ on the plane is assigned to a unique coordinate $(P_x, P_y)$.

We know that the equation of the circle with center $O$ and radius $r$ is $$(x - O_x)^2 + (y - O_y)^2 - r^2 = 0.$$

The formula for the power of a point is ultimately related to the above equation of the circle $$\vec{PU} \times \vec{PV} = PO^2 - r^2 = (P_x - O_x)^2 + (P_y - O_y)^2 - r^2.$$

This formula $$(P_x - O_x)^2 + (P_y - O_y)^2 - r^2$$ signifies the reason why the value of the power of a point is

*positive*,

*zero*, and

*negative*when the point is

*outside*the circle,

*on*the circle, and

*inside*the circle.

Today we have learned some basic facts about the power of a point to a circle. In the next post, we will learn more about its properties and its many applications.

Power of a point: $\vec{PU} \times \vec{PV} = PO^2 - r^2 = (P_x - O_x)^2 + (P_y - O_y)^2 - r^2$. |

Before we conclude the post, let us summarize the three basic formulas for the power of a point: $$\vec{PU} \times \vec{PV} = PO^2 - r^2 = (P_x - O_x)^2 + (P_y - O_y)^2 - r^2.$$

Hope to see you again next time.

*Homework.*

1. Suppose that the point $P$ is

*outside*the circle $(O)$. Draw a tangent $PT$ to the circle. Apply Pythagorean Theorem to show that $$PU \times PV = PT^2 = PO^2 - r^2.$$

2. Suppose that the point $P$ is

*inside*the circle $(O)$. Draw the line $UV$ perpendicular to $PO$. Again, apply Pythagorean Theorem to show that $$PU \times PV = PU^2 = r^2 - PO^2.$$

3. Given two circles $(O_1)$ and $(O_2)$ intersecting at two points $I$ and $J$. Show that for any point $P$ on the line $IJ$, $P$ has equal power with respect to these two circles.

4. Given two circles $(O_1)$ and $(O_2)$. Determine the locus of all the points that have equal power to these two circles.

5. Given three circles $(O_1)$, $(O_2)$ and $(O_3)$. Determine the point $P$ such that it has equal power to all of these circles.

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