$$C:x^2+y^2=r^2$$ $$A(0,A_y)$$
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?