Find range of values for p in equation of circle

Question

Can somebody please check my working with the following question:

Given the equation ${x^2 + y^2 - 2px - 4py + 3p + 2 = 0}$ represents a circle, determine a range of values for p.

I don't think I can use the discriminant because there are y values so I can use:

${g^2 + f^2 - c > 0}$

g = -p

f = -2p

c = 3p + 2

${(-p)^2 + (-2p)^2 - 3p + 2 > 0}$

=> ${p^2 +4p^2 -3p + 2 > 0}$

=> ${5p^2 -3p + 2 > 0}$

I thought I would then find values for p and display them like ${p_1 < p < p_2}$

I was going to use the quadratic formula:

=> ${{3 \pm \sqrt{(-3)^2 - 4(5)(2)}}\over 5}$

But the discriminant is a negative number so I don't think I am on the right path.

Answer 1

It should be $-3p-2$ rather than just $-3p+2$ and you are on right path.

Answer 2

HINT:

$$(x-p)^2+(y-2p)^2=5p^2-3p-2=(5p+2)(p-1)$$ which needs to be positive

Can you take it home from here?