Equations of Tangent and Normal at a Point on a Given Circle

Diameter form (Theorem 5.2). Let $A(x_1,y_1)$ and $B(x_2,y_2)$ be the two ends of a diameter, and $P(x,y)$ any point on the circle. Since an angle in a semicircle is a right angle, $\angle APB=90^\circ$, so the chords $AP$ and $PB$ are perpendicular and the product of their slopes is $-1$:

the equation of the circle with $A,B$ as the ends of a diameter.

Theorem 5.3 (position of a point relative to a circle). For the circle $x^2+y^2+2gx+2fy+c=0$ with centre $C(-g,-f)$ and radius $r$, and a point $P_1(x_1,y_1)$: draw $CP_1$, meeting the circle at $Q$. Then $P_1$ is outside/on/inside the circle according as $CP_1>,=,<CQ(=r)$, i.e. according as

So simply substituting the point's coordinates into the circle's expression (call this value $S_1$) and reading its sign tells you instantly where the point lies — no distance computation needed.

Tangent and normal at a point on the circle. For $P(x_1,y_1)$ and $Q(x_2,y_2)$ both on $x^2+y^2+2gx+2fy+c=0$, subtracting their two equations and simplifying gives the slope of chord $PQ$ as $-\dfrac{(x_1+x_2)+2g}{(y_1+y_2)+2f}$. Letting $Q\to P$ turns the chord into the tangent at $P$, with slope $-\dfrac{x_1+g}{y_1+f}$; substituting this slope into the point-slope form and simplifying (using that $(x_1,y_1)$ itself satisfies the circle's equation) gives the clean result

The normal, perpendicular to the tangent at the same point, has slope $\dfrac{y_1+f}{x_1+g}$ and simplifies to …