Answer
See explanations.
Work Step by Step
Step 1. Given $x^2+y^2=a^2$, we have $2xdx+2ydy=0$ and $\frac{dy}{dx}=-\frac{x}{y}$ which gives the slope of tangent lines to the circle.
Step 2. For a line tangent to the circle at point $(x_0,y_0)$. its slope is $m=-\frac{x_0}{y_0}$ and the slope of its normal is $n=-1/m=\frac{y_0}{x_0}$
Step 3. The equation for the normal line can be written as $y-y_0=\frac{y_0}{x_0}(x-x_0)$ which leads to $y=\frac{y_0}{x_0}x$ and clearly this line passes through the origin.
