Two tangents of the circle $x^2+y^2=8$ at $A$ and $B$ meet at $P =(-4,0)$ then find the area of quadrilateral $PAOB$.
Here, $O$ is the origin.
Distance of $P$ from $O$ is $4$ unit.
$$OA=OB=2\sqrt 2$$
Therefore $$OP^2=OA^2+AP^2$$ $$AP=2\sqrt 2$$
Hence $$\Delta POA=4$$ $$ar(PAOB)=8$$
This was a true of false question. I modified it a bit. Basically, the answer says the area is not 8sq unit. What am I doing wrong.
There is another sub question to this
Is the area of quadrilateral formed by the tangents from an external point with length of tangent 2r?
From the way I have solved, I am inclined to say no, but since it says my answer is wrong, please help me verify this.