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

Possible values are 4 and - 4.

Step-by-step explanation:

First complete the square on the x and y terms:

x^2 + y^2 + px + 6y - 3 = 0

(x + 0.5p) ^2 - 0.25p^2 + (y + 3) ^2 - 9 - 3 = 0

(x + 0.5p) ^2 + (y + 3) ^2 = 12 + 0.25p^2

This is the standard form of the equation of the circle where the right side = radius^2 so we can write:

12 + 0.25p^2 = 4^2

0.25p^2 = 16 - 12 = 4

Taking square roots:-

0.5p = + / - 2

p = 2 / 0.5, - 2 / 0.5

p = 4, - 4.

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.