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!
Hint:
If those curves are tangent to the curve, their slope is given by $\;y'=3x^2+1\;$ at each general point $\;(x,\,x^3+x)\;$ on the curve , so for what points $\;(a,\,a^3+a)\;$ on the curve are there lines through them and through $\;(2,2)\;$ whose slope is $\;3a^2+1\;$ ?
Best Answer
Since the two pairs of tangents are symmetric with respect to the $y$-axe, the quadratic function $f(x)=ax^{2}+bx+c$ must be even ($f(x)=f(-x)$), which implies that $b=0$. The equations of the tangents to the graph of $f(x)=ax^{2}+c$ at points $% \left( x_{1},f(x_{1})\right) $ and $\left( x_{2},f(x_{2})\right) $ are $$\begin{eqnarray*} y &=&f^{\prime }(x_{i})x-f^{\prime }(x_{i})x_{i}+f(x_{i})\qquad i=1,2 \\ &=&2ax_{i}x+c-ax_{i}^{2}. \end{eqnarray*}$$
These equations must be equivalent to two of the given tangents, one from each pair, e.g. $y=2x-10$ and $y=x-4$:
$$\left\{ \begin{array}{c} 2ax_{1}x+c-ax_{1}^{2}=2x-10 \\ 2ax_{2}x+c-ax_{2}^{2}=x-4% \end{array}% \right. $$
Finally we compare coefficients and solve the resulting system of $4$ equations:
$$\left\{ \begin{array}{c} 2ax_{1}=2 \\ c-ax_{1}^{2}=-10 \\ 2ax_{2}=1 \\ c-ax_{2}^{2}=-4% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} x_{1}=8 \\ x_{2}=4 \\ a=\frac{1}{8} \\ c=-2% \end{array}% \right. $$
Thus the quadratic is $f(x)=\frac{1}{8}x^{2}-2$.