Find the equations of the two lines which pass through the point $(0,4)$ and form tangents to a circle of radius $2$, centered on the origin.
Firstly, we have the equation of the circle $x^{2}+y^{2}=4$. Which we can rearrange to get $y$ in terms of $x$:
$$y^{2}=4-x^{2}\implies y=\pm\sqrt{4-x^{2}}$$
However, we know that the tangents must touch points on the top half of the circle, therefore we can simply take the principle square root, $y=\sqrt{4-x^{2}}$. Moreover, since the lines originate from the same point, and are tangential to the same circle, the two tangential points are $(x,\sqrt{4-x^{2}})$ and $(-x,\sqrt{4-x^{2}})$.
The gradients of the tangents at these points can be found by implictly differentiating the original equation and obtaining:
$$\frac{dy}{dx}=\frac{-x}{y}=\left\{\frac{-x}{\sqrt{4-x^{2}}},\frac{x}{\sqrt{4-x^{2}}}\right\}$$
However, I'm unsure how to go about completing this problem. I know it's a simple question, but I simply cannot see how to solve it.
Thanks in advance!
Best Answer
Call $\,P=(0,4)\,$ , and let $\,A:=(a,b)\,$ be one of the two tangency points. Since the tangent line's perpendicular to the circle's radius at the point of tangency, we get:
$$m_{AP}=-\frac{b-4}{a}\,\,,\,m_A=\frac{b}{a}\Longrightarrow \frac{b}{a}=\frac{a}{b-4}\Longrightarrow a^2=b^2-4b$$
But we also have $\,a^2+b^2=4\Longrightarrow a^2=4-b^2\,$ , since $\,A\,$ belongs to the circle, so we get the quadratic
$$a^2=4-b^2=b^2-4b\Longrightarrow b^2-2b-2=0\Longrightarrow b_{1,2}=1\pm\sqrt2$$
and now find out the $\,a's\,$ and etc...and without calculus!