Finding all tangent lines to circle $x^2+y^2=9$ having exactly one common point with parabola $y=x^2+3$ 
I have to find all tangent lines to the circle $x^2+y^2=9$, which have exactly one common point with the parabola $y=x^2+3$.

I drew it and I saw that $(0,3)$ is a common point circle and parabola. I found that three tangent lines are: $y=3$, $x=-3$, $x=3$. And now the question is whether these are all? I don't know how to explain it or find others lines.
Thanks for your help!

At any given point on the circle, there is a unique tangent line.  At the point $(0,3)$ you have identified 3.  Only one is tangent.
The other lines:

Since this is tagged pre-calculus, I will not use calculus in this answer.
Suppose $x_1,y_1$ lies on the circle.
Slope of the tangent line though that point is negative reciprocal of the slope of the radius.
$m = -\frac {x_1}{y_1}$
Equations of the tangent line in point slope and y-intercept form:
$y-y_1 = -\frac {x_1}{y_1} (x-x_1)\\
y = -\frac {x_1}{y_1} x + \frac {x_1^2 +y_1^2}{y_1}\\
y = -\frac {x_1}{y_1} x + \frac {9}{y_1}\\
$
Remember $x_1,y_1$ lie on the circle so $x_1^2 + y_1^2 = 9$  we will be using this fact a couple more times.
The line is also tangent to $y = x^2 + 3$
Substituting:
$-\frac {x_1}{y_1} x + \frac {9}{y_1} = x^2 + 3\\
y_1x^2 + x_1x + 3y_1 - 9 = 0$
Since our line is tangent, when we plug this into the quadratic formula, we must get exactly one root.  i.e. $b^2 - 4ac = 0$.
$x_1^2 - 4y_1(3y_1-9) = 0\\
x_1^2 + 36y_1 - 12y_1^2 = 0$
We can use $x_1^2+ y_1^2 = 9$ again. Substitute $x_1^2 = 9 - y_1^2$
$(9-y_1^2) + 36y_1 - 12y_1^2 = 0\\
13y_1^2 - 36y_1 - 9 = 0\\
(y_1 - 3)(13y_1 + 3) = 0$
We have two values for $y_1.$  Solve for $x_1.$
$x_1^2 + 9 = 9\\
x_1^2 + \frac {9}{169} = 9$
$(0,3),(\frac {3\sqrt {168}}{13},-\frac {3}{13}), (-\frac {3\sqrt {168}}{13},-\frac {3}{13})$ are our 3 points of tangency.
And, plug into our earlier equations of the tangent lines.
$y = 3\\      
y = \sqrt {168} x - 39\\
y = -\sqrt {168} x - 39$
