long double is always printed 0

Question

I wrote a program that calculates the square root of a floating-point number using long double precision.

#include <stdio.h>
#include <tgmath.h>

int main(void)
{
    long double x;
    printf("Enter a real number: ");
    scanf("%Lf", &x);
    //x = 2.0; //Not a scanf() problem, even with this line it still outputs 0
    int precision;
    printf("Enter the precision of the output (number of digits after the .): ");
    scanf("%d", &precision);
    printf("sqrt(%.*Lf) = %.*Lf\n", precision, x, precision, sqrt(x));
    //Tried with a constant precision embedded in the string instead of the *, still prints 0
    printf("Press Enter to continue . . . ");
    getchar();
    getchar();
}

The output is:

Enter a real number: 2
Enter the precision of the output (number of digits after the .): 5
sqrt(0.00000) = 0.00000
Press Enter to continue . . .

At first, I though it was a bug with my program. I replaced the long double with double and %Lf with %lf and it all seems to work:

#include <stdio.h>
#include <tgmath.h>

int main(void)
{
    double x;
    printf("Enter a real number: ");
    scanf("%lf", &x);
    int precision;
    printf("Enter the precision of the output (number of digits after the .): ");
    scanf("%d", &precision);
    printf("sqrt(%.*f) = %.*f\n", precision, x, precision, sqrt(x));
    printf("Press Enter to continue . . . ");
    getchar();
    getchar();
}

Output:

Enter a real number: 2
Enter the precision of the output (number of digits after the .): 5
sqrt(2.00000) = 1.41421
Press Enter to continue . . .

What is the reason for this?

Compiler: GCC 9.2.0 (with -std=c17)

Platform: Windows 10 x64

UPDATE: A simple printf("%Lf\n", 2.0L); prints 0.000000. This means my code is not bugged.