Circle and right triangle

Question

I'm considering $A=(-1,0)$, $B=(1,0)$ and the unit circle of center $O=(0,0)$. The dot $K=(x,y)$ is on the circle. My aim is to prove that $ABK$ is a right triangle. Since $AB=2KO$, I have $$ AB^2=4x^2+4y^2.$$ Moreover, $BK^2=(x-1)^2+y^2$ and $AK^2=(x+1)^2+y^2$. Hence, $BK^2+AK^2=2x^2+2y^2+2$. How can I prove simply , without using circle equation ect, that $$ AB^2=AK^2+KB^2\quad ?$$ Did I miss an information?

Answer 1

I get $$AB=(2,0),AK=(x+1,y),BK=(x-1,y)$$ So $$4=(x+1)^2+y^2+(x-1)^2+y^2$$ This is $$4=2x^2+2y^2+2$$ Since we have the unit circle $$x^2+y^2=1$$ we get $$4=2+2$$. That is true.

Answer 2

Relate what you have found so far to the numbers that were given. Since $AB$ is a diameter from $(-1,0)$ to $(1,0)$ it has length $2$. Combining this with your expression for $AB$, or by simply noting that $K$ is a point on the unit circle, we have $x^2+y^2=1\Rightarrow y=\sqrt{1-x^2}$. Then what are the sizes of $AK^2$ and $BK^2$?