Answer
See below for detailed work.
Work Step by Step
$$x^2+y^2=a^2$$
1) Find the derivative of $y$, using implicit differentiation: $$2x+2yy'=0$$ $$x+yy'=0$$ $$y'=-\frac{x}{y}$$
2) As the tangent and the normal at a point is perpendicular with each other, the product of their slopes equals $-1$.
So if we call the slope of the tangent $y'$ and that of the normal $s$, we have $$s\times y'=-1$$ $$s=-\frac{1}{y'}=-\frac{1}{-\frac{x}{y}}=\frac{y}{x}$$
At a point $A(x_0,y_0)$ on the circle, we have $s(x_0)=\frac{y_0}{x_0}$.
The normal to the circle at $A$ is $$y-y_0=\frac{y_0}{x_0}(x-x_0)$$ $$y-y_0=\frac{y_0}{x_0}x-y_0$$ $$y=\frac{y_0}{x_0}x$$
The form of this equation shows that the normal line to the circle at any point $A$ always passes through the origin.
