Question

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

Answer 1

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.

Answer 2

Possible values are 4 and -4.

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.