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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 $$(1, 1)$$ for :

JEE · Math · previous-year question

  1. A.all except one value of $$p$$correct
  2. B.all except two values of $$p$$
  3. C.exactly one value of $$p$$
  4. D.all values of $$p$$

Answer

A. all except one value of $$p$$

Explanation

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 $${S_1} - {S_2} = 0$$ $$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$ $$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is $${S_1} + \lambda \,\,L = 0$$ $$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda $$ $$\left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$ 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 $$p=-1$$

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