<s>[INST] 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 $y=-6x$.
 [/INST] 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
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