It does not exist because the left and right do not match
Here, the given equation is,
x^2+y^2+px+6y-3=0
By copairing it with x^2+y^2+2gx+2fy+c=0, we get,
2g=p or, g=p/2
2f=6 or, f=3
c=-3
now the radius is,
r^2=g^2+f^2-c
or, 4^2=(p/2)^2 +3^2 -(-3)
or, 16= p^2/4 +9+3
or, 16-9-3=p^2/4
or p^2/4=4
or, p^2=4×4
or p^2=4^2
or p=4
therefore the rewuired value of p is 4.
