1. Abridged Problem Statement  
-----------------------------  
Given a disc in the plane of radius r centered at (x₀, y₀), with the origin strictly outside the disc. The density at any point (x,y) is p(x,y)=ln(x²+y²). Compute the total mass  
   M = ∬_{disc} ln(x²+y²) dx dy  
and print it with a relative error up to 10⁻¹².

2. Detailed Editorial  
---------------------  
Let s = √(x₀² + y₀²). Since the origin is outside the disc, s > r. We want  
  M = ∬_{(x–x₀)²+(y–y₀)² ≤ r²} ln(x²+y²) dx dy.  

Step A: Change to polar coordinates (D,θ) centered at (x₀,y₀):  
  x = x₀ + D cosθ,  
  y = y₀ + D sinθ,  
  0 ≤ D ≤ r,  0 ≤ θ < 2π,  
  dx dy = D dD dθ.  

Step B: Express x²+y²:  
  x²+y² = s² + D² + 2 s D cosθ.  

So  
  M = ∫_{θ=0 to 2π} ∫_{D=0 to r} D · ln(s² + D² + 2sD cosθ)  dD dθ.  

Step C: For fixed D (0≤D≤r), let A = s² + D², B = 2sD.  One uses the known integral  
  I(A,B) = ∫₀^{2π} ln(A + B cosθ) dθ  
when A > |B|; the result is  
  I = 2π · ln( (A + √(A² – B²)) / 2 ).  
Here  
  √(A² – B²) = √((s² + D²)² – 4s²D²) = s² – D²,  
  so (A + √(A² – B²))/2 = (s² + D² + s² – D²)/2 = s².  
Thus  
  ∫₀^{2π} ln(s² + D² + 2sD cosθ) dθ = 2π·ln(s²).  

Step D: Then  
  M = ∫₀^r D · [2π ln(s²)] dD = 2π ln(s²) ∫₀^r D dD  
    = 2π ln(s²) · (r²/2) = π r² ln(s²)  
    = π r² · 2 ln(s) = π r² · ln(s²).  

Hence the final formula:  
  M = π r² · ln(x₀² + y₀²).

3. Provided C++ Solution with Line-by-Line Comments  
---------------------------------------------------  
#include <bits/stdc++.h>  
using namespace std;  

// Overload operator<< for pairs, to allow easy printing  
template<typename T1, typename T2>  
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {  
    return out << x.first << ' ' << x.second;  
}  

// Overload operator>> for pairs, to allow easy input  
template<typename T1, typename T2>  
istream& operator>>(istream& in, pair<T1, T2>& x) {  
    return in >> x.first >> x.second;  
}  

// Overload operator>> for vectors, to read a sequence of elements  
template<typename T>  
istream& operator>>(istream& in, vector<T>& a) {  
    for (auto& x : a) in >> x;  
    return in;  
}  

// Overload operator<< for vectors, to print elements separated by space  
template<typename T>  
ostream& operator<<(ostream& out, const vector<T>& a) {  
    for (auto x : a) out << x << ' ';  
    return out;  
}  

const double pi = acos(-1);  // Value of π  

double x, y, r;  // Global variables for input  

// Read the three input numbers into x, y, r  
void read() {  
    cin >> x >> y >> r;  
}  

void solve() {  
    // Compute s² = x² + y²  
    // The derived formula for the mass is π * r² * ln(s²).  
    double ans = pi * r * r * log(x * x + y * y);  

    // Print with fixed format and 12 digits after decimal  
    cout << setprecision(12) << fixed << ans << endl;  
}  

int main() {  
    ios_base::sync_with_stdio(false);  
    cin.tie(nullptr);  

    int T = 1;           // Single test case  
    for (int test = 1; test <= T; test++) {  
        read();          // input x, y, r  
        solve();         // compute and print the answer  
    }  
    return 0;  
}  

4. Python Solution with Detailed Comments  
-----------------------------------------  
```python
import math
import sys

def main():
    # Read three integers from standard input
    data = sys.stdin.read().split()
    x0, y0, r = map(int, data)

    # Compute squared distance from origin to center
    s2 = x0*x0 + y0*y0

    # According to the closed-form result:
    # mass = π * r² * ln(s²)
    mass = math.pi * (r**2) * math.log(s2)

    # Print with high precision
    print("{:.12f}".format(mass))

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
-----------------------  
Mass M = ∬_{disc} ln(x²+y²) dx dy.  
Shift to polar coords around (x₀,y₀); integrate θ using ∫₀²π ln(A + B cosθ)dθ = 2π ln((A+√(A²–B²))/2). With A = s²+D², B = 2sD, one gets 2π ln(s²). Then ∫₀ʳ D dD = r²/2. Final M = π r² ln(x₀²+y₀²).