The normal form of a circle's equation is $x \cos \alpha + y \sin \alpha = p$. Here, 'p' represents the length of the perpendicular drawn from the origin (0,0) to the tangent line, and '$\alpha$' is the angle that this perpendicular makes with the positive x-axis, measured counterclockwise. This form is derived by considering a tangent line to the circle. The shortest distance from the origin to this tangent is 'p', and the direction of this shortest distance is given by the angle '$\alpha$'. Imagine a circle centered at the origin. If we consider a tangent line, the line segment from the origin to the point of tangency is perpendicular to the tangent. The length of this segment is the radius. However, the normal form is about the distance from the origin to any line, not necessarily a tangent to a circle centered at the origin. For a general line $Ax + By + C = 0$, the perpendicular distance from the origin is $\frac{|C|}{\sqrt{A^2 + B^2}}$. If we normalize this equation such that the constant term is positive and the coefficient of the perpendicular distance is 1, we get the normal form. Specifically, if we divide $Ax + By + C = 0$ by $-\sqrt{A^2 + B^2}$ (assuming C is positive, or adjust sign accordingly), we get $\frac{-A}{\sqrt{A^2 + B^2}}x + \frac{-B}{\sqrt{A^2 + B^2}}y = \frac{-C}{\sqrt{A^2 + B^2}}$. Let $\cos \alpha = \frac{-A}{\sqrt{A^2 + B^2}}$ and $\sin \alpha = \frac{-B}{\sqrt{A^2 + B^2}}$ (adjusting signs based on quadrant of $\alpha$), and $p = \frac{C}{\sqrt{A^2 + B^2}}$. This yields $x \cos \alpha + y \sin \alpha = p$.