Circle and tangents from an external point (I)

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

• (a) draw it;
• (b) find its equation;
• (c) draw the two tangent lines to the circle through the point $P(-5;0)$ and find their equations.

Solution

(a)–(b) $(x-3)^2+(y+3)^2=9\ \Longleftrightarrow\ x^2+y^2-6x+6y+9=0$.

(c) Pencil of lines through $P(-5;0)$: $y=m(x+5)$, i.e. $mx-y+5m=0$. Imposing distance from $C(3;-3)$ equal to $3$: $\frac{|3m-(-3)+5m|}{\sqrt{m^2+1}}=\frac{|8m+3|}{\sqrt{m^2+1}}=3\Rightarrow (8m+3)^2=9(m^2+1).$ $55m^2+48m=0\Rightarrow m(55m+48)=0\Rightarrow m=0\ \vee\ m=-\tfrac{48}{55}.$ Tangents: $y=0$ (the $x$-axis) and $y=-\dfrac{48}{55}(x+5)=-\dfrac{48}{55}x-\dfrac{48}{11}$.

$\boxed{x^2+y^2-6x+6y+9=0;\quad y=0,\ \ y=-\tfrac{48}{55}x-\tfrac{48}{11}}$