Tangents to a circle by several methods

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Consider the circle with centre $A(-2;-1)$ and radius $3$.

• (a) write its equation;
• (b) find the tangents from $P(3;2)$ (distance method);
• (c) find the tangents parallel to $y=-2x$ (distance method);
• (d) find the tangent at the point of the circle with $x=-1$ (radius-perpendicular method).

Solution

(a) $(x+2)^2+(y+1)^2=9$, i.e. $x^2+y^2+4x+2y-4=0$.

(b) Line $y-2=m(x-3)$. Distance from $A$ equal to $3$: $\dfrac{|{-5m}+3|}{\sqrt{m^2+1}}=3\Rightarrow 16m^2-30m=0\Rightarrow m=0\ \vee\ m=\tfrac{15}{8}$. Tangents $y=2$ and $y=\tfrac{15}{8}x-\tfrac{29}{8}$.

(c) Line $2x+y-q=0$. $\dfrac{|-5-q|}{\sqrt5}=3\Rightarrow q=-5\pm3\sqrt5$. Tangents $y=-2x-5\pm3\sqrt5$.

(d) The point with $x=-1$: $y=-1\pm2\sqrt2$. At $T\left(-1;-1+2\sqrt2\right)$ the radius has slope $2\sqrt2$, so the tangent has slope $-\tfrac{\sqrt2}{4}$: $y=-\tfrac{\sqrt2}{4}(x+1)-1+2\sqrt2$. Similarly the other point gives slope $+\tfrac{\sqrt2}{4}$.

$\boxed{\text{(b) } y=2,\ y=\tfrac{15}{8}x-\tfrac{29}{8};\quad \text{(c) } y=-2x-5\pm3\sqrt5}$