Find the equation of the tangents given the gradient

Question

We're given that $x^2 + (y+2)^2 = 4$ and we're asked to find the equation of the lines where the gradient $=1$

Through implicit differentiation I got $x + (y+2)y'=0$ and if $y'=1$ then:

$y=-x-2$ is the relation where $y'=1$, which makes no sense to me.

I'm missing on something, so how do you go about finding the equations? (I know there're two.)

Answer 1

By implicit differentiation of $x^2+(y+2)^2=4$, we get $$2x+2(y+2)y'=0$$ or $$y'=\frac{-x}{y+2}$$ If we isolate $y$ in the equation of the circle, we have $$y=\sqrt{4-x^2}-2$$ hence $$y'=\frac{-x}{\sqrt{4-x^2}-2+2}$$ $$=\frac{-x}{\sqrt{4-x^2}}$$ Now letting $y'=1$ and solving for $x$ yields the two solutions $x=\pm \sqrt{2}$. These give the corresponding $y$-values of $\mp \sqrt{2}-2$.

Using point-slope form, the two equations are then $$y+\sqrt{2}+2=x-\sqrt{2}$$ and $$y-\sqrt{2}+2=x+\sqrt{2}$$ or in more commonly-used slope-intercept form as $$y=x-2\sqrt{2}-2$$ $$y=x+2\sqrt{2}-2$$

Answer 2

the equation $x^2 + (y+2)^2 = 4$ is the graph of the circle centered at $(0, -2)$ and of radius $2.$ tangents at the end of the diameter joining the points $(\sqrt 2, -2 - \sqrt 2), (-\sqrt 2, -2 + \sqrt 2)$ has the slope $1.$ they are given by $$x - y = \pm(2\sqrt 2 - 2).$$