Answer
Normal on a circle passes through the centre of the circle. This can proved using the equation of a normal at an arbitrary point on a circle.
Work Step by Step
Equation of a circle is given by $$x^2+y^2=r^2$$
Slope at any point $(x,y)$ on the circle is obtained by doing implicit differentiation with respect to $x$ on the circle equation.
$$2x + 2yy'=0$$
$$slope = y'=-\frac{x}{y}$$
This slope is also the slope of tangent to the circle at $(x, y)$. Slope of normal is the negative reciprocal of the slope of tangent. Thus, normal slope $n$, $$n=\frac{y}{x}$$
Thus, the equation of normal at $(x_1, y_1)$ is,
$$y = \frac{y_1}{x_1}x$$
$$\implies x_1y = y_1x$$
The center of the above circle is $(0,0)$. This point satisfies the equation of the normal. Thus, normal passes through the centre of the circle.
