The radius of the circle x^2+y^2 +px+6y-3=0 is 4. find the possible values of p

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.

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