One possibility is for each quadratic to share a root.

Multiplying (x-p)(x-q) and (x-q)(x-r) give equations for c in terms of (p,q,r) with solutions c=3 and c=-1.

The only other possibility is for one quadratic to have a repeated root.  

Say the first quadratic has the dual root p.  Since the discriminant is postive, p is real.  But it's necessary too for p^2=-4(c^2+1) which is impossible.

If it's the second quadratic with the repeated root then 2p=4 and p^2=-2c(c^2+1).  So p=2 and substituting gives a cubic equation with a real root c=-1 by inspection and the remaining roots imaginary.

When c=3 (p,q,r)=(10,-4,-6).

When c=-1 (p,q,r)=(2,1+sqrt(17),1-sqrt(17)).