Tangent line problems Problem 1
Find common tangent to the curve:
$y+x^2=-4$ and $x^2+y^2=4$.
My idea:

Let $t1... y=ax+b$ is a tangent line to the first curve.
Let $t2... y=cx+d$ is a tangent line to the second curve.
But we are seracing for common tangent so that mean that $a=c$ and $b=d$.
By using implicit differentiation we can find $a$ and $b$.
$$y+x^2=-4$$
$$y'+2x=0$$
$$y'=-2x$$
$$x^2+y^2=4$$
$$2x+2yy'=0$$
$$x+yy'=0$$
$$y'=\frac{-x}{y}$$
$$-2x=\frac{-x}{y} \to y=\frac{1}{2},$$$x$ is any real number
Problem 2
For which value of the coefficients $a$, $b$ and $c$ $\in$ R is the x-coordinate axis tangent to the curve?
$y=ax^2+bx+c$
My idea:
 Curve and the tangent line must have just one common point. Our curve is a parabola and the number of dots on x axis is a number of solution of quadratic equation. So we need $a,b,c$ to be equal to $b^2-4ac=0$ 
Problem 3
Find a line that is tangent to the curve
$y = x^4 - 2x^3 - 3x^2 + 5x + 6$
in at least two points .
My idea:
Let that tangent be $t=ax+b$.
If we that line be a tangent line at two point to the curve equation $y(x)-t(x)=0$ must have at least two (not equal) solution.
$$x^4 - 2x^3 - 3x^2 + 5x + 6-(ax+b)=0$$
$$x^4 - 2x^3 - 3x^2 + 5x + 6=ax+b$$
$$a=5 , b=6$$
$$x^4 - 2x^3 - 3x^2=0$$ $$x_1=0, x_2=-1, x_3=3$$
Problem 4
I am trying to find the number of tangents to a curve that all pass through the origin. The curve's equation is $y=x^3+x^2−22x+20.$ I also need to find the equation of said tangents.
My work:
Let's use formula for tangent line:
$$y-y_0=y'(x_0)(x-x_0)$$
We know that $x=0$ and $y=0$.
Let's find $y'(x_0)$:
$y'(x_0)=3x_0^2+2x_0-22$ 
Plugin what we know:
$$-y_0=(3x_0^2+2x_0-22)*-x_0$$
We aslo know that $y_0=x_0^3+x_0^2-22x_0+20$
$$-x_0^3-x_0^2+22x_0-20=-3x_0^3-2x_0^2+22x_0$$
$$2x_0^3+x_0^2-20=0$$
Only real solution is 2. $x_0=2$ and $y_0=-12$.
Let tangent line be $y=kx+l$.
$k=y'(2)=-6$
$-12=-6*2+l\to l=0$
But we want l to be zero. 
Our solution is $y=-6x$.
 A: For problem 1, note that the requirement is that two points(not necessarily the same) must share a tangent line(which means both $a=c$ and $b=d$). That they might not be the same point means you cannot assign them the same x and y values. So we have to simultaneously solve:
$$-2x_1=x_2/y_2$$
And:
$$b=d$$
To find expressions for b and d, we can use:
$$y=y_k'(x-x_k)-y_k$$
That is, $b=x_1y_1'-y_1$ and $d=x_2y_2'-y_2$.
You have to simultaneously solve:
$$-2t=-u/\sqrt{4-u^2}$$
And:
$$-t^2+4=-u^2/\sqrt{4-u^2}-\sqrt{4-u^2}$$
Wolfram is giving me a double surd solution to these equations. This seems pretty nasty to solve. Have you got the questions right?
Your answer to problem 2 is correct and well reasoned. =]
For problem 3, you need to find a line $y=ax+b$ such that $P(x)-y(x)=(x-\alpha)^2(x-\beta)^2$
$$x^4-2x^3-3^2+(5-a)x+(6-b)\equiv(x-\alpha)^2(x-\beta)^2$$
Expanding the right hand side:
$$x^4-2x^3-3^2+(5-a)x+(6-b)\equiv x^4-2(\alpha+\beta)x^3+(\alpha^2+\beta^2+4\alpha\beta)x^2-2(\alpha^2\beta+\alpha\beta^2)x+\alpha^2\beta^2$$
We can find $\alpha$ and $\beta$ by solving:
$$-2(\alpha+\beta)=-2$$
And:
$$\alpha^2+\beta^2+4\alpha\beta=-3$$
Giving us $\alpha=1$ and $\beta=-2$.
From there:
$$x^4-2x^3-3x^2+(5-a)x+(6-b)\equiv x^4-2x^3-3x^2+4x+4$$
That is, $5-a=4\to a=1$ and $6-b=4\to x=2$. Our tangent is $y=x+2$
A: Another solution for problem 3:
Let solve the more general problem:

Let $p(x)$ be a polynomial of degree $n\ge 4$. Find bitangents to the
  curve $y=p(x)$.

Let $y-p(x_0)=p'(x_0)(x-x_0)$ be the equation of the tangent at the point $(x_0,p(x_0))$. It will be a bitangent if the abscissae equation for the intersection points of this line with the curve has another double root.
We'll write this equation using the Taylor's expansion of $p(x)$ at $x=x_0$:
$$p(x)=p(x_0)+p'(x_0)(x-x_0)+\frac{p''(x_0)}{2}(x-x_0)^2+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^n.$$
The equation for the intersection points is thus
$$\frac{p''(x_0)}{2}(x-x_0)^2+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^n$$
and the other intersection points abscissae are roots of
$$\frac{p''(x_0)}{2}+\frac{p'''(x_0)}{6}(x-x_0)+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^{n-2}$$
So the abscissae of the other points of contact with the curve are the multiple roots of this equation which are different from $x_0$.
In the present case,  as $n=4$, you get a quadratic equation, which must have a double root in $X=x-x_0$:
$$6x_0^2-6x_0-1+(4x_0-3)X+X^2=0,$$
whence the condition
$$(4x_0-3)^2-4(6x_0^2-6x_0-1)=-8x_0^2+13=0.$$
A: Problem$\#1:$
The equation of tangent  of $$x^2+y+4=0$$ at $(t,-(4+t^2)),$ 
$$x(t)+\dfrac{y+\{-(4+t^2)\}}2+4=0\iff2t x+y+4-t^2=0$$
Now this will be a tangent of $$x^2+y^2=4$$ if the perpendicular distance of tangent to center of circle $=$ radius of circle
