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