flat approximation of a circle at a point

Question

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.

what do you menas exactly for ''flat''? ifyou menas linear approximation use the tangent line. — Emilio Novati, Mar 17 '15 at 19:25
ok. That makes sense. I think I forgot how to do some stuff from calculus. — user2809114, Mar 17 '15 at 19:33
You need to evaluate the derivative at $x = 1$ to get the numerical slope of the tangent line. :) — Andrew D. Hwang, Mar 17 '15 at 19:35

Emilio Novati · Answer 1 · 2015-03-17T19:52:14.530

1

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: $$ y-\dfrac{\sqrt{5}}{2}=-\dfrac{2\sqrt{5}}{5}(x-1) $$

edited Mar 17 '15 at 19:52

answered Mar 17 '15 at 19:39

Emilio Novati

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flat approximation of a circle at a point

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