Example — tangents with the delta method

Example — Tangents from an external point, the $\Delta$ method

Find the tangents to the circle $\gamma:\ x^2 + y^2 - 2x - 4y + 4 = 0$ passing through $P(0;4)$.

Centre and radius: $C(1;2)$, with $R^2 = \dfrac{(-2)^2}{4} + \dfrac{(-4)^2}{4} - 4 = 1 + 4 - 4 = 1$, that is $R = 1$.

Pencil through $P$: $y - 4 = m\,x$, that is $mx - y + 4 = 0$. We set up the system with $\gamma$: $\begin{cases} y = mx + 4 \\ x^2 + y^2 - 2x - 4y + 4 = 0 \end{cases}$ Substituting $y$ into the second equation: $x^2 + (mx+4)^2 - 2x - 4(mx+4) + 4 = 0 \implies (1+m^2)x^2 + (4m-2)x + 4 = 0.$

Tangency: $\Delta = 0$. $\Delta = (4m-2)^2 - 16(1+m^2) = 16m^2 - 16m + 4 - 16 - 16m^2 = -16m - 12.$ $\Delta = 0 \iff m = -\tfrac{3}{4}$. From the pencil we find a single tangent.

The excluded generator. The pencil $y - 4 = m\,x$ does not include the vertical line $x = 0$. Let us check it directly: substituting $x = 0$ into the equation of $\gamma$ gives $y^2 - 4y + 4 = 0 \iff (y-2)^2 = 0 \iff y = 2$ (double root). Hence $x = 0$ meets $\gamma$ at a single point with multiplicity two: it is tangent.

Solution: two tangents, $\boxed{y = -\tfrac{3}{4}x + 4 \ \text{e}\ x = 0.}$ ✓

The two tangents to $\gamma$ drawn from the external point $P$: the line $y=-\tfrac34 x+4$ and the vertical line $x=0$.

Links

Topics: Circonferenza analitica
Concepts: Discriminante · Fascio di rette · Punto esterno · Retta · Retta tangente
Methods: Circonferenza tangenti
Skills: Geometria analitica · Risolvere equazioni