Question

how close does the semicircle

y=radical(16-x^2) come to the point

(1,radical3)?

Answer 1

let P(a,b) be the closest point.
the slope of the tangent at P is -a/b

then the slope of the line from P to (1,√3) must be b/a, since that line must be at right angles to the tangent for a minimum distance.

but slope of line from P to (1,√3) = (b-√3)/(a-1)

so (b-√3)/(a-1) = b/a
.
.
b = a√3
sub this into y=√(16-x^2) and squaring both sides gave me
a = 2

so the closest point is (2,2√3)
and the closest distance is

√[(2-1)^2 + (2√3 - √3)^2]

= √(1 + 3) = 2

Answer 2

Wait, why are you substituting what b equals into the original equation... How does that even help you in the first place?