If $AB$ is the diameter of a circle and $P$ another point on the circumference, Euclidean geometry tells us that angle $APB = 90˚$. Use this fact to show that the equation of a circle whose diameter has endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.
I have tried doing it using the Midpoint and Radius but got stuck in the middle of algebra. Is that the correct way? How do I use the Euclidean fact?
Best Answer
Let $P(x,y)$ be any point on the circle. It's easily shown that
$$\vec{AP}=(x-x_1,y-y_1),\vec{BP}=(x-x_2,y-y_2). $$
Since $\angle APB=\dfrac{\pi}{2}$, we know that $\vec{AP}\cdot\vec{BP}=0$, which is the same as
$$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0. $$
Geometrically, one finds $k_{AP}\cdot k_{BP}=-1$ when $\angle APB=\dfrac{\pi}{2}$, so
$$\dfrac{x-x_1}{y-y_1}\cdot\dfrac{x-x_2}{y-y_2}=-1, $$
which can be simplified to the preceding result.