The equation of a line with slope m that passes through the point (a,b) is:
[tex]y=m(x-a)+b[/tex]On the other hand, if a curve is defined by a function y=f(x), the line tangent to the curve at a point (a,f(a)) will have a slope equal to f'(a), so the equation of the line tangent to the curve at (a,f(a)) is given by the expression:
[tex]y=f^{\prime}(a)(x-a)+f(a)[/tex]Isolate y from the given equation to find an expression for f(x):
[tex]\begin{gathered} (x+2)^2+y^2=4 \\ \\ \Rightarrow y^2=4-(x+2)^2 \\ \\ \Rightarrow y=\pm\sqrt{4-(x+2)^2} \end{gathered}[/tex]A function must have only 1 value for each input. In this case, y has two possible values for each given value of x. We want to explore the x-intercepts, which are given by the condition y=0. Then, any of the two signs is useful for this purpose. We will choose the positive sign. Then:
[tex]f(x)=\sqrt{4-(x+2)^2}[/tex]Find the derivative of f:
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}f(x) \\ =\frac{d}{dx}\sqrt{4-(x+2)^2} \\ =\frac{1}{2\sqrt{4-(x+2)^2}}\cdot\frac{d}{dx}\left(4-(x+2)^2\right) \\ =\frac{1}{2\sqrt{4-(x+2)^2}}\cdot-2(x+2) \\ =-\frac{x+2}{\sqrt{4-(x+2)^2}} \\ \\ \therefore\quad f^{\prime}(x)=-\frac{x+2}{\sqrt{4-(x+2)^2}} \end{gathered}[/tex]Find the values of x that correspond to y=0:
[tex]\begin{gathered} \left(x+2\right)^2+y^2=4 \\ \Rightarrow\left(x+2\right)^2+0=4 \\ \Rightarrow\left(x+2\right)^2=4 \\ \Rightarrow x+2=\pm\sqrt{4} \\ \Rightarrow x+2=\pm2 \\ \Rightarrow x=-2\pm2 \\ \Rightarrow x_1=-4,x_2=0 \end{gathered}[/tex]The x-intercepts are -4 and 0, but the denominator of the derivative is equal to 0 when x reaches any of those values, so the derivative diverges.
Then, the slope of the line tangent to the circle at the x-intercepts is not defined, which means that those lines are vertical lines.
Therefore, the equation of the lines tangent to the given circle at the x-intercepts, are:
[tex]\begin{gathered} x=-4 \\ x=0 \end{gathered}[/tex]
