Equation of a tangent line for circles

Question

When calculating a tangent to a circle, is the method the same as tangent to a curve?

Problem: A circle has a radius of $2$ and is centered at the origin. Find the the equation of the tangent line to the circle at $(1, \sqrt{3})$.

From here, would I just isolate $y$ and use implicit differentiation, and then plug $1$ into that - or is the correct method different from that?

Answer 1

This seems like a needless calculus approach. We know that the tangent line is perpendicular to the radius. You just need a vector so that

$[a,b]\cdot [1,\sqrt{3}]=0$. In other words, $[\sqrt{3},-1]$ will suffice. So the line is just parametrized by: $(1,\sqrt{3})+t[\sqrt{3},-1]$.

the slope is also clear this way: $\frac{-1}{\sqrt{3}}$...

If you haven't seen vectors, this is equivalent to taking the so-called "negative reciprocal"

If you insist on a calculus approach, at least use polar co-ordinates so you don't have to take an annoying derivative.

Answer 2

a=2; Differentiate $ \sqrt{a^2-x^2} $

or implicitly find slope from

$$ 2 x + 2 y \frac{dy}{dx} =0 $$

and then use straight line (point, slope) form for tangent.

Same for any curve incl. circle.