Tangents with the distance method

Problem

Find the tangents to $\gamma:\ x^2 + y^2 - 4x - 2y - 4 = 0$ passing through $P(5;5)$ with the distance method.

Solution

Centre and radius: $a=-4$, $b=-2$, $c=-4$, hence $C(2;1)$ and $R^2 = 4 + 1 + 4 = 9$, that is $R = 3$.

Pencil through $P$: $y - 5 = m(x - 5)$, that is $mx - y + 5 - 5m = 0$ (the vertical line $x = 5$ is excluded from the pencil and must be checked separately).

Distance equal to the radius: $d(C,r) = \frac{|m\cdot 2 - 1 + 5 - 5m|}{\sqrt{m^2+1}} = \frac{|4 - 3m|}{\sqrt{m^2+1}} = 3.$ Squaring: $(4-3m)^2 = 9(m^2+1) \iff 16 - 24m + 9m^2 = 9m^2 + 9 \iff 24m = 7 \iff m = \tfrac{7}{24}.$

The pencil gives a single tangent: $y - 5 = \tfrac{7}{24}(x-5)$, that is $y = \tfrac{7}{24}x + \tfrac{85}{24}$.

The excluded vertical. We substitute $x = 5$ into $\gamma$: $25 + y^2 - 20 - 2y - 4 = 0 \iff y^2 - 2y + 1 = 0 \iff (y-1)^2 = 0$, double root. Hence $x = 5$ is tangent.

Solution: two tangents, $\boxed{x = 5 \ \text{e}\ y = \tfrac{7}{24}x + \tfrac{85}{24}.}$

Links

Topics: Circonferenza analitica
Concepts: Distanza punto retta · Fascio di rette · Punto esterno · Retta · Retta tangente
Methods: Circonferenza tangenti
Skills: Geometria analitica · Risolvere equazioni
Exercise types: Problema geometrico