Perpendicular property of tangents: $\bf w \bullet u$ $= r^2$

Question

Points $A$ and $C$ with position vectors $\bf u$ and $\bf v$ lie on the circle centre $\bf 0$ and radius $r$. Tangents to the circle at $A$ and $C$ meet at the point $B$ with position vector $\bf w$.

How can I show, using the perpendicular property of tangents, that $\bf w \bullet u$ $= r^2$ and $\bf w \bullet v$ $= r^2$ and that $\bf \vec{OB}$ is perpendicular to $\bf \vec{AC}$?

Answer 1

Calling $\;U,\,V\;$ the tangents at $\;A,C\;$ corr., we get by adding geometrically vectors that:

$$w=u+U\implies w\cdot u=(u+U)\cdot u=u\cdot u+u\cdot U=r^2+0=r^2$$

and the same follows for $\;w\cdot v\;$ .

Since $\;\vec{AC}=-u+v\;$ , we get:

$$\vec{OB}\cdot\vec{AC}=w\cdot(-u+v)=-w\cdot u+w\cdot v=-r^2+r^2=0\implies\vec{OB}\perp\vec{AC}$$