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Power of a point to a circle

Today we will learn about the 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.

1 comment:

1. wonderful! Thanks a lot