If $$P$$ and $$Q$$ are the points of intersection of the circles $${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$       and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$       then there is a circle passing through $$P, \,Q$$  and $$\left( {1,\,1} \right)$$  for :

Solution :
The given circles are
$$\eqalign{ & {S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0.....(1) \cr & {S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0.....(2) \cr} $$
$$\therefore $$ Equation of common chord $$PQ$$  is $${S_1} - {S_2} = 0$$
$$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$
$$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is $${S_1} + \lambda \,{\text{L}} = 0$$
$$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda \left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$
As it passes through (1, 1), therefore
$$ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$
$$ \Rightarrow \lambda = - \frac{{2p + 7}}{{{{\left( {p + 1} \right)}^2}}},$$       which does not exist for $$p=-1$$