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.

circle area = pi * radius^2

if diameter = 10 then radius = 5

circle area = pi * 5^2

circle area = pi * 25

circle area = 78.5398163397 square units

step-by-step explanation:

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.