I need to find a flat approximation of a circle at a given point. The circle I am working with is $$x^2+y^2=\frac9 4$$
The point is $(1,\sqrt{\frac 5 4})$
I have found an approximation, but it is not flat. Here is what I did:
$$\left( y = \sqrt{\frac 9 4 -x^2})\right)\lor \left( y = -\sqrt{\frac 9 4 -x^2})\right) $$
The derivative of the top half of the graph gives me $$ y = - \frac{x}{\sqrt{ \frac 9 4}- x^2} $$
Using function transformation, I got $$ y = - \frac{x-1}{\sqrt{\frac 9 4 - (x-1)^2}} +\sqrt{\frac 5 4} $$
Which looks like a good approximation, but again, it is not flat.
Hint:
Note that your circle is centered at the origin. The tanget at a point is orthogonal to the radius so if the coordinate of the point are $(1, \sqrt{5/4})$ the slope of the tangent is $-\dfrac{2\sqrt{5}}{5}$ and the equation of the tangent line is: $$ y-\dfrac{\sqrt{5}}{2}=-\dfrac{2\sqrt{5}}{5}(x-1) $$