Where's the error in my calculation of a line through a point and being the tangent to a circle? $$C:x^2+y^2=r^2$$
$$A(0,A_y)$$
I'd like to find the line L through A and being a tangent on C.
Define point P on C.
$$P(P_x,P_y)$$
$$P_x^2+P_y^2=r^2$$
Get the slope of L, by calculating the derivative in P
$$x^2+y^2=r^2 \Rightarrow f(x)=y=\sqrt{r^2-x^2}$$
$$ {f(x) \over dx} = {-1 \over 2 \sqrt{r^2-x^2}} (-2x) = {x \over {\sqrt{r^2-x^2}}}$$
$$ {f(P_x) \over dx} = {P_x \over {\sqrt{r^2-P_x^2}}} $$
Use point A and the slope to put together an equation defining L
$$ L:y-A_y={P_x \over {\sqrt{r^2-P_x^2}}}(x-0)$$
Insert point P
$$ P_y-A_y={P_x \over {\sqrt{r^2-P_x^2}}}(P_x-0)$$
$$ P_y-A_y={P_x^2 \over {\sqrt{r^2-P_x^2}}}$$
Replace Px
$$ P_x^2 = r^2-P_y^2 $$
$$ P_y-A_y={r^2-P_y^2 \over P_y}$$
$$ 2P_y^2-A_yP_y-r^2=0 $$
Now the problem here is that this equation's form is not correct. With the center of Circle C being(0, 0), and point A being on the Y axis, I expect an equation of the form:
$$ P_y^2-k^2=0 $$
Does anyone see any error in this calculation?
 A: Note that you can write the tangent line at the point $(x,y)$ as $(x,y)+t(-y,x)$.
So you want to solve 
$$(x,y)+t(-y,x)=(0,A_y)$$
under the condition $x^2+y^2=r^2$.
The first equation gives you $x=ty$ and $y(1+t^2)=A_y$. Can you finish from there?
Btw: You will get two solutions, where both have the same $y$ but different $x$. Do you see why this happens geometrically?
A: Two suggestions: stick with simple notation; if you get stuck with an explicit derivative (which you didn't), try implicit differentiation.
Here they are, side by side (the explicit formula on the LHS
technically needs a $\pm$ sign, which I omit because I'm not using it further):
$$ y=f(x)=\sqrt{r^2-x^2}\qquad\stackrel{\pm y}{\iff}\qquad x^2+y^2=r^2 $$
$$ y'=\frac12\left(r^2-x^2\right)^{-1/2}\cdot(-2x)\qquad\qquad 2x+2yy'=0 $$
$$ y'=-x\left(r^2-x^2\right)^{-1/2}\qquad\qquad yy'=-x $$
$$ y'=-\frac{x}{\sqrt{r^2-x^2}}\qquad\qquad y'=-\frac{x}y $$
Now if you are given a point $A=(0,a)$ on the $y$ axis
and wish to determine the point $P=(x,y)$
where this intersects the circle at a tangent,
you can exploit the geometry of the situation
to derive the equation
$$
\frac{y-a}x\cdot\frac{y}x=-1
$$
since $OP \perp AP$ (and perpendicular lines have negative reciprocal slopes), where the origin $O=(0,0)$ is the center of the circle. So you can try solving this together with the equation of the circle (above right). Your system of equations is then:
$$ x^2+y^2=r^2 $$
$$ x^2+y(y-a)=0 $$
Would you agree that it seems simpler to solve?
