Finding x when slope = 1 I've been working out some problems relating to slope on the points of a curve. I'm having issues with this one: In the curve to which the equation is... $$x^2 + y^2 = 4$$
find the value of $x$ at those points where the slope $= 1$. I thought to differentiate and then set $\frac{dy}{dx}$ equal to $1$ and solve for $x$. This seems right but even differentiating this particular equation was confusing to me. As always, many thanks in advance for any help on this. 
 A: A simple Hint:
When $F(x,y)=0$ defines $y$ as a function of $x$ implicitly, as above, then we have $$y'=\frac{-F_x}{F_y}$$ in which $F_x$ means the partial differential of $F$ with respect to $x$. 


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*Note that if you want to use this way, you should make sure that the relation $F(x,y)=0$ make a functional relation between $y$ and $x$.

A: HINT:
Applying derivative, $$2x+2y\frac{dy}{dx}=0\implies \frac{dy}{dx}=-\frac xy$$
$$\implies -\frac xy=1\iff y=-x $$
Put the value of $y$ in the given equation and solve the resulting Quadratic Equation of $x$.

Alternatively using coordinate Geometry,  the equation of the tangent will be $y=1\cdot x+c$ 
Let us find the intersection of the given circle with the starlight line.
Putting the value of $y$ in the given equation
$$(x+c)^2+x^2=4\iff 2x^2+2c x+c^2-4=0$$ which is Quadratic Equation in $x$
For tangency, both the root must be same as the point of intersection must coincide.
So, the discriminant must $(2c)^2-4\cdot2(c^2-4)=0\implies 4c^2=32\implies c=\pm2
\sqrt2$
Consequently, $x=-\frac{2c}2=-c$
