The parametric points on a circle for its diameter

Question

Suppose we have a circle $x^2+y^2=a^2$

Any point of the circle will be $(a\cos\theta,a\sin\theta)$

Why does the diameter through this point intersect the circle at $(-a\cos\theta,-a\sin\theta)$?

I think it’s weirdly obvious, and I also have an idea on why it might be so, but I don’t have a proper derivation. It’s probably very simple, but I just can’t work it out.

Answer 1

Well, the center of the circle at $(0,0)$ divides the diameter into two parts of equal lengths $a$. Hence using the midpoint formula, we get

$$X:{m+a\cos\theta\over 2}=0$$ $$Y:{n+a\sin\theta\over 2}=0$$

Clearly, the other end of the diameter $(m,n)$ is $(-a\cos\theta,-a\sin\theta)$

Answer 2

The centre of this circle is the origin. The origin is also the mid-point of any pairs of points $(x,y)$ and $(-x,-y)$.

Does this answer your question?

(Note that you have missed out one of the $a$s in your second point.)