Prove this property for an arbitrary circle Prove that in an arbitrary circle, the point on the circle closest to the origin must lie on the extended line connecting the circle's centre and the origin.


Let the equation of the circle be $(x-a)^2+(y-b)^2=r^2$
So, any point on the circle can be $P(r\cos \theta+a,r\sin \theta+b)$
If the distance$(d)$ from the point $(r\cos \theta+a,r\sin \theta+b)$ from the origin,
$d^2=(r\cos \theta+a)^2+(r\sin \theta+b)^2=r^2+a^2+b^2+2r(a\cos \theta+b\sin \theta)=r^2+a^2+b^2+2r\sqrt{a^2+b^2}\cos(\theta-\arctan\frac b a)$
This will be minimum if $\cos(\theta-\arctan\frac b a)=-1$
$\implies \theta-\arctan\frac b a=(2n+1)\pi$ where $n$ is any integer.
So, $\tan \theta=\tan\{(2n+1)\pi+\arctan\frac b a\}=\frac b a\implies b\cos \theta=a\sin \theta$
Now the area of the triangle with vertices $(0,0),(a,b),(r\cos \theta+a,r\sin \theta+b)$ is
$$\frac 1 2\det\begin{pmatrix} 0 & 0 & 1 \\ r\cos \theta+a & r\sin \theta+b & 1 \\ a & b&1\end{pmatrix}=\frac 12 \left(b(r\cos \theta+a)-a(r\sin \theta+b)\right)=\frac{r(b\cos \theta-a\sin \theta)}2=0$$
So,  $(0,0),(a,b),(r\cos \theta+a,r\sin \theta+b)$ are co-linear for the minimum distance of $P$ from the origin $O(0,0)$ for any $a,b,r$.
A little generalization:
If the distance of $P(r\cos \theta+a,r\sin \theta+b)$ from $Q(c,d)$ is $D,$
$D^2=(r\cos \theta+a-c)^2+(r\sin \theta+b-d)^2$
$=r^2+(a-c)^2+(b-d)^2+2r\{(a-c)\cos \theta+(b-d)\sin \theta\}$
Applying the same method,the distance will be minimum is $(b-d)\cos \theta=(a-c)\sin \theta$
Now the area of the triangle with vertices $(c,d),(a,b),(r\cos \theta+a,r\sin \theta+b)$ is
$$\frac 1 2\det\begin{pmatrix} c & d & 1 \\ r\cos \theta+a & r\sin \theta+b & 1 \\ a & b&1\end{pmatrix}$$
$$=\frac 12\det\begin{pmatrix} c-a & d-b & 1-1 \\ r\cos \theta+a-a & r\sin \theta+b-b & 1-1 \\ a & b&1\end{pmatrix}=\frac{r\{(a-c)\sin\theta-(d-b)\cos\theta\}}2=0$$
