The Knight's estate

Question

And then the King said: You fought bravely, Knight, and your deed will not be forgotten for centuries. For your valor I grant you this castle and the lands around it. Things rush me, and I can not take you to the castle. Therefore, I will give you the way from this place to the castle. Now go and come back after the deadline. - as it is written in the Green Book of Years.

In addition, it is known from the Green Book of Years that the lands with which the castle was granted were in the shape of a circle. The king was very wise and, in order to avoid unnecessary proceedings regarding the right to land, always granted only areas of land on the map that have a convex shape.

Recently, historians have had information about where the castle was located and where this historical conversation took place. They want to know how much land did the Knight get on the assumption that the road to the castle was perfectly straight.

Explanation

The following figure shows in light gray the territory originally granted to the knight, and in dark gray, the one that came to him as a result of the king giving him the way.

Input

The first line of the input contains two floating-point numbers: x_k and y_k - the coordinates of the place where the dialogue took place. The second line contains three floating-point numbers: x_c, y_c and r_c - the coordinates of the castle and the radius of the circle that bounds the land granted with it.

Output

Print one floating-point number - the area of ​​the land obtained by the Knight, with an accuracy of at least three characters after the decimal point.

Tests

Input    Output

2 5       5.69646
2 1 1

---

3 9       80.7130
2 3 5

---

1 3       3.141
1 2 1

Note: A triangle may not include the entire semicircle if it is too close to the center, as in the test I have given.

Answer 1

Wolfram Language (Mathematica), 51 bytes

Area@ConvexHullMesh[CirclePoints[##2,7!]~Append~#]&

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                    CirclePoints[##2,7!]            (* generate 5040 points on the perimeter of the circle *)
                                        ~Append~#   (* append the point at the end of the way, *)
Area@ConvexHullMesh[                             ]& (* and find the area of the convex hull of those points *)

Answer 2

JavaScript (ES7), 76 bytes

Port of Jitse's answer.

with(Math)f=(X,Y,x,y,r,h=hypot(X-x,Y-y))=>r*(r*(PI-acos(r/h))+(h*h-r*r)**.5)

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Answer 3

Python 3, 107 bytes

from math import*
def f(X,Y,x,y,r):d=hypot(X-x,Y-y);return(r>d)*pi*r*r or(pi-acos(r/d))*r*r+(d*d-r*r)**.5*r

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-3 bytes thanks to Arnauld

-4 bytes thanks to squid

-5 bytes thanks to Greg Martin

Explanation

First, we calculate the traveled distance between both coordinates using the Pythagorean theorem. Let's call it d.

Then, we construct a rectangular triangle between coordinate 1, coordinate 2 and the point on the circle where it meets with the straight line. Using the Pythaogrean theorem again, the area of this triangle is given by $$\frac{1}{2}\times\sqrt{d^2-r^2}\times r$$ where d is the distance between the coordinates and r is the radius of the circle. Since we have two of those triangles, this area is resembled in the code by (d*d-r*r)**.5*r

Next, we need to calculate the remaining area of the circle. A circle's area is given by $$ a = \pi\times r^2$$ However, part of the circle is already taken into account. The angular fraction of the circle that is already described by the triangles is given as $$\cos^{-1}(r/d)$$ so that the remaining area in the circle can be described by (pi-acos(r/d))*r*r.

And then some code juggle to make it as short as possible.

Except...

when the traveled distance lies within the circle. Then just return the circle area.

Answer 4

Jelly, 33 bytes

½÷@ÆAØP_ײ}
I²SḢ_²}½×ʋ+çɗṛ²×ØPʋ>?

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Port of @jitse’s Python answer so be sure to upvote that one! A full program taking as its arguments [[xk, xc], [yk, yx]] and r.

Answer 5

Python 3, 96 bytes

from math import*
def f(X,Y,x,y,r):t=max(hypot(X-x,Y-y)**2/r/r-1,0)**.5;return(t-atan(t)+pi)*r*r

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In Python 3.8, the solution can be further improved using assignment expressions:

Python 3.8 (pre-release), 91 bytes

lambda X,Y,x,y,r:((t:=max(hypot(X-x,Y-y)**2/r/r-1,0)**.5)-atan(t)+pi)*r*r
from math import*

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-2 bytes thanks to @Jitse

This solution is based on a bit more math manipulation of the formula. Let \$\theta\$ be the size of common angle of each triangle and its corresponding sector. The area of each triangle equals to \$\frac{1}{2}\times r\tan{\theta}\times r\$ and the area of the common sector of the triangles and the circle equals to \$\theta\times r\times r\$. Hence, the total area $$S=\tan{\theta}\cdot r^2 +\pi r^2-\theta r^2=(\pi+\tan{\theta}-\theta)r^2$$

We define \$t\$ to be $$t=\tan{\theta}=\frac{\sqrt{d^2-r^2}}{r}=\sqrt{\frac{d^2}{r^2}-1}$$ Here, \$d\$ is the distance between the two given coordinate pairs. Then we have $$S=(\pi+t-\arctan{t})r^2$$

Answer 6

C (clang), 111 103 93 bytes

float f(X,x,r)_Complex X,x,r;{X=cabs(X-x)/r;return(4*atan(1)-atan(X=csqrt(X*X-1))+X)*r*r;}

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It's 90 bytes for the code, but +3 bytes because we need to pass -lm to the compiler.

Thanks to @ceilingcat for shaving off 12 bytes, and the brilliant use of csqrt() inspired me to take the coordinates as complex numbers, to shave off another 10 bytes.