Circle and tangents from an external point (II)

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Consider the circle with centre $C(-2;2)$ and radius $3$.

• (a) draw it;
• (b) find its equation;
• (c) draw the two tangent lines to the circle through the point $P(3;3)$ and find their equations.

Solution

(a)–(b) $(x+2)^2+(y-2)^2=9\ \Longleftrightarrow\ x^2+y^2+4x-4y-1=0$.

(c) Pencil of lines through $P(3;3)$: $y-3=m(x-3)$, i.e. $mx-y+3-3m=0$. Imposing distance from $C(-2;2)$ equal to $3$: $\frac{|-2m-2+3-3m|}{\sqrt{m^2+1}}=\frac{|1-5m|}{\sqrt{m^2+1}}=3\Rightarrow (1-5m)^2=9(m^2+1).$ $16m^2-10m-8=0\ \Rightarrow\ 8m^2-5m-4=0\ \Rightarrow\ m=\frac{5\pm3\sqrt{17}}{16}.$ Hence $m_1\approx 1.086$ and $m_2\approx -0.461$. Tangents $y-3=m(x-3)$ with these two slopes.

$\boxed{x^2+y^2+4x-4y-1=0;\quad m=\tfrac{5\pm3\sqrt{17}}{16}}$