Let S be the set of all points A on the circle $x^ 2 + (y − 2)^2 = 1$ so that the tangent line at A has a non-negative y-intercept; then S is the union of one or more circular arcs. Find the total length of S
From the equation, we have the center at $(0,2)$
I let $(a,b)$ be the point that satisfies the conditions
then
$a^2 + (b-2)^2 =1$
Equation of the tangent line on the line from $(0,2)$ and $(a,b)$
= $\frac{a}{2-b}(x-a) = y-b$
= $\frac{ax}{2-b} + \frac{a^2}{b-2}+b = y$
y Intercept must be non-negative
therefore,
$\frac{a^2+b^2-2b}{b-2} > 0$
Using the Circle Equation.
=$\frac{2b-3}{b-2} > 0$
therefore, [3/2, 2] are the bounds, which has 15 degrees, but the answer is not 23/12 pi. I don't know where I went wrong, and I think what I did was reasonable, I am also asking for solutions that are shorter than this one. Thanks!