If P and Q are the points of intersection of the circles x2+y2+3x+7y+2p−5=0 and x2+y2+2x+2y−p2=0, then there is a circle passing through P,Q and (1,1) for:
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Official solution
The given circles are
\[{S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\]
\[{S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\]
\[\therefore\] Equation of common chord PQ is S1−S2=0
\[ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0\]
\[ \Rightarrow \] Equation of circle passing through P and Q is
\[{S_1} + \lambda \,\,L = 0\]
\[ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda \]
\[\left( {x + 5y + {p^2} + 2p - 5} \right) = 0\]
As it passes through \[\left( {1,1} \right),\] therefore
\[ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0\]
\[ \Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},\]
which does not exist for p=−1