Finding the equation of a tangent of a circle at a point

Question

The line with equation y=mx is tangent to the circle with centre (-8,0) and radius 4 at the point P(x,y)

Show that $m=\pm\frac{\sqrt{3}}{3}$ and hence find the coordinates of P

Answer 1

$y = mx$ is the tangent line, so the equation $(x+8)^2 + (mx)^2 = 4^2$ has only $1$ real solution. So: $(1 + m^2)x^2 + 16x + 48 = 0$ must have: $\triangle' = 8^2 - 48(1+m^2) = 0$, or $64 - 48 - 48m^2 = 0$ or $16(1 - 3m^2) = 0$ or $ 1 - 3m^2 = 0$ or $m = \dfrac{1}{\sqrt{3}}$ or $m = -\dfrac{1}{\sqrt{3}}$.

So: $x = -\dfrac{8}{1+m^2} = -6$, and $y = mx = 2\sqrt{3}$, and $y = -2\sqrt{3}$. So:

$P = (-6,2\sqrt{3})$, and $(-6,-2\sqrt{3})$