পাওয়ার অফ পয়েন্ট প্রকাশের একটি ভিন্ন পদ্ধতি

In the diagram, $O$ is the center of the circle, which has (let's say) radius $R$. Out of the 3 ways to represent the power above, the third way ($PA \cdot PB$) is the least useful, since $A, B$ don't have much information attached to them, since they're just random points on the circle. Let's use the definition of power equals $PT^2$ first.

Note that since $PT$ is a tangent to the circle, $\angle PTO = 90^\circ$. Therefore, by the Pythagorean Theorem, $PT^2 = OP^2-OT^2 = \boxed{OP^2-R^2}$.

Now let's use the $PR \cdot PQ$ definition. Then the power is $PR \cdot PQ = (OP-OR)(OP+OQ) = (OP-R)(OP+R) = \boxed{OP^2-R^2}$, again. Therefore, we can also define the power of a point $P$ in the plane with respect to a circle with center $O$ and radius $R$ to be $OP^2-R^2$.

Now, let's look at the equation of a circle. The equation of a circle is $(x-a)^2+(y-b)^2 = R^2$, which can also be written as $(x-a)^2+(y-b)^2-R^2 = 0$. Now, let the point $P$ have coordinates $(x_1, y_1)$.

When we plug in the point $P$ for $x, y$ on the LHS of the above equation, what do we get? We get $(x_1-a)^2+(y_1-b)^2-R^2$. But what does $(x_1-a)^2+(y_1-b)^2$ actually mean? Well, it's just the distance from $(x_1, y_1)$ to $(a, b)$ (the center), squared, so it actually just equals $OP^2$!. So $(x_1-a)^2+(y_1-b)^2-R^2 = \boxed{OP^2-R^2}$, and we once again get the power of a point!