how close does the semicircle

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

(1,radical3)?

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

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

To find out how close the semicircle y = √(16 - x^2) comes to the point (1, √3), we can follow these steps:

Step 1: Calculate the distance between the center of the circle and the given point.
- The center of the semicircle is at the point (0, 0) because the equation is in the form y = √(r^2 - x^2), which represents a circle centered at the origin.
- The distance between (0, 0) and (1, √3) can be calculated using the distance formula:

d = √[(x2 - x1)^2 + (y2 - y1)^2]
d = √[(1 - 0)^2 + (√3 - 0)^2]
d = √[1 + 3]
d = √4
d = 2

Step 2: Subtract the radius of the semicircle from the distance obtained in Step 1.
- The radius of the semicircle is √16, which is 4.
- Subtracting the radius from the distance gives us:

2 - 4 = -2

-2 represents the distance between the semicircle and the given point (1, √3). However, since distances cannot be negative, we can conclude that the semicircle does not come close to the point (1, √3); it lies 2 units away in the opposite direction.

To determine how close the semicircle given by the equation y = √(16-x^2) gets to the point (1, √3), we need to find the distance between the point and the curve. Here's how you can calculate it step-by-step:

1. Start with the equation of the semicircle: y = √(16 - x^2).

2. Substitute the x-coordinate of the point into the equation to determine the corresponding y-coordinate. In this case, x = 1. Plug it in: y = √(16 - 1^2) = √15.

3. Calculate the distance between the two points using the distance formula: √[(x2 - x1)^2 + (y2 - y1)^2]. In this case, (x1, y1) = (1, √3) and (x2, y2) = (1, √15). Therefore, the distance formula becomes:
√[(1 - 1)^2 + (√15 - √3)^2] = √(0 + 12) = √12.

4. Simplify the square root of 12. Since 12 can be factored into 2 * 2 * 3, √12 = √(2^2 * 3) = 2√3.

So, the semicircle y = √(16 - x^2) comes as close as 2√3 units away from the point (1, √3).