Find the equation of a tangent line at $(3,-1)$ on the circle $x^2+y^2+2x-y-17=o$

Question

Determine the equations of tangents from point $A( 3 , -1)$ to circle (C) of equation: $x^2+y^2+4x+8y+3=0$ Thanks in advance :)

Answer 1

Hints:

$$x^2+4x+y^2+8y+3=0\implies (x+2)^2+(y+4)^2=17$$

Let $\;(a,b)\;$ one of the tangency points on the circle, then if the wanted line is

$$y+1=m(x-3)\iff y=mx-3m-1\;\;\text{(why?)}$$

it must be that

$$m\cdot\frac35=-1\iff m=-\frac53\;\;\text{(why?)} $$

Answer 2

$$x^2+y^2+4x+8y+3=0$$ Apply the derivative w.r.t. x to both sides: $$2x+2y\frac{dy}{dx}+4+8\frac{dy}{dx}=0$$ Now we make $\frac{dy}{dx}$ the subject of the formula: $$\dfrac{dy}{dx}(2y+8)=-2x-4$$ $$\dfrac{dy}{dx}=\dfrac{-2x-4}{2y+8}$$ $$\dfrac{dy}{dx}=-\dfrac{x+2}{y+4}$$ Now we insert $(x,y)=(3,-1)$ and find the slope of the tangent line $$\dfrac{dy}{dx}=-\dfrac{5}{3}$$ Now that we have the slope, we can find the equation of the line as we know it passes through the point $$y-y_1=m(x-x_1)$$ $$y-1=-\dfrac{5}{3}(x-3)$$ $$y-1=-\dfrac{5x-15}{3}$$ $$y-1=-\dfrac{5x}{3}+5$$ $$y=-\dfrac{5x}{3}+6$$ And theres the equation for your line!