<s>[INST] Equation of a tangent on a circle given the gradient and equation of the circle My maths teacher told me this problem was impossible without knowledge of implicit differentiation: is she right?
You are given the equation of the circle $\left(x+2\right)^2+\left(y-2\right)^2=16$ , what are the equations of the lines with a gradient of 2 which are tangents of this circle.
Here is a visualisation of the problem 
The red circle is the one I described, and the green and blue lines are the equations I'm looking for. 
I think the solution will have something to do with plugging in the value of 2 in some sort of equation for the gradient i.e. $2=\frac{x-x_0}{y-y_0}$ , and then the discriminant the equation you would use to find where they intersect can be solved to find values of $x$ , but I'm not sure how to find these values without using any implicit differentiation -can it be done?
EDIT:
Ok so now I'll try and run this through step by step until i find another problem:
So we can say that there's a line (the diameter) of gradient -1/2 which passes through the centre (-2,2):
$y=mx+c$ 
$y=-\frac{1}{2}x+c$
$2=1+c$
$y=-\frac{1}{2}x+1$
We can also say that a solution lies on the point $(a,b)$ such that
$\left(a+2\right)^2+\left(b-2\right)^2=16$ and $b=-\frac{1}{2}a+1$
So we can substitute b in the equation of the circle to arrive at
$\left(a+2\right)^2+\left(-\frac{1}{2}a-1\right)^2=16$
which simplifies down to
$\frac{5}{4}a^2+5a-11=0$
and solves to give $\frac{2}{5}\left(-5-4\sqrt{5}\right)$ and $\frac{2}{5}\left(4\sqrt{5}-5\right)$
so these are the possible  values for $x$ 
Now if i substitute 1 of these back into the equation, I'll get two values of $y$ for the one value of $x$ because it's quadratic, but there's only 1 intersection point at each value of $x$ , what do i do at this point?
EDIT 2:
It looks like both the intersection points of the line $x=\frac{2}{5}\left(4\sqrt{5}-5\right)$  correspond to the $y$ values of the intersection points of the tangents, does this mean i only need to plug one of these $x$ values back into the equation? 
Visualisation to show you what i mean (the blue line appears to intersect the circle at both the appropriate $y$ values)

 [/INST] $$ (x+2)^2 + (y-2)^2 = 16 $$
$$ \Rightarrow x^2 + y^2+4x-4y+4+4 = 16 $$
$$ \Rightarrow x^2 + y^2+4x-4y-8= 0 $$
$$ \text{centre} =(-2,2), \text{ radius}=4 $$
 for tangent the line must satisfies the condition, $r=d$ 
we have line $y=2x+c$ here gradient$=2$ need to find  $c=$?
$$ \pm4 =\frac {2(-2)-2+c}{2^2+1} $$
$$ \pm 20 =-6 +c$$
$$ c= 26,-14$$
now the tangents are
$$y=2x+26, y=2x-14$$ are required tangents 
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