Find tangents to a circle parallel to the straight line

Question

Find tangents to a circle $x^2+y^2=5$ parallel to the straight line $2x-y+1=0$

My solution:

$$x^2+y^2=5$$ $$S=(0,0)$$ $$r=\sqrt{5}$$ $$y=2x-1$$ $$a=2$$

Searching for b $$y=2x+b$$

Using following formula: $$d=\frac{Ax_0+By_0+C}{\sqrt{A^2+B^2}}$$

$$-2x+y-b$$ $$\sqrt{5}=\frac{-2\cdot0+1\cdot0+b}{\sqrt{(-2)^2+1^2}}$$ $$\sqrt{5}=\frac{b}{\sqrt{5}}$$

$b= 5$ or $b= -5$

Tangents: $$y=2x+5$$ $$y=2x-5$$

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Seems like it should be right since we got $\sqrt{5}$ at the end, but could someone look into it to make sure?

Answer 1

1. The correct distance formula between a point $P_0(x_0,y_0)$ and the straight line $Ax+Bx+C=0$ is $$d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}.\tag{1}$$
2. From $(1)$ and since the center of the given cirle ( $ x^2+y^2=5 $ ) is $(x_0,y_0)=(0,0)$, we obtain $$d=\frac{|C|}{\sqrt{A^2+B^2}}.\tag{2}$$
3. The straight line $2x−y+1=0$ (thick blue, in the figure below) is perpendicular to the straight line $(1/2)x+y=0$ (red), because two straight lines with equations $$Ax+By+C=0\qquad\text{ and } \qquad A'x+B'y+C'=0\tag{3}$$ are perpendicular if and only if $$AA'+BB'=0.\tag{4}$$ This line $(1/2)x+y=0$ defines the two tangent points to the circle, $(2,-1)$ and $(-2,1)$, which are the two solutions of the system of simultaneous equations, one representing the given circle $x^2+y^2=5$ (thick black) and the other, the straight line $(1/2)x+y=0$: $$ \left\{ \begin{aligned} x^2+y^2 &=5 \\ \frac{1}{2}x+y &=0 \end{aligned} \right. \tag{5} $$
4. The family of the parallel lines to the given line $2x-y+1=0$ is defined by equation $$2x−y+K=0,\tag{6} $$ where $K$ is a parameter.
5. The two tangent lines, whose equations have been found by you ( $y=2x\pm 5$ ) correspond to $K=\pm 5$.

$$\mathbf{Figure.}\text{ Circle } x^2+y^2=5, \text{given line (thick blue) } 2x−y+1=0, \text{and the two tangents at the points } (2,−1) \text{ and }(−2,1) \text{ (thin blue).} $$