Q43 / 90 · JEE Main 2009 · Mathematics

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

  • Correct answer
    A
    all values of $p$
  • B
    all except one value of $p$
  • C
    all except two values of $p$
  • D
    exactly one value of $p$

Solution

Given circles $S=x^2+y^2+3 x+7 y+2 p-5=0$ $$ S^{\prime}=x^2+y^2+2 x+2 y-p^2=0 $$ Equation of required circle is $S+\lambda S^{\prime}=0$ As it passes through $(1,1)$ the value of $\lambda=\frac{-(7+2 p)}{\left(6-p^2\right)}$ If $7+2 p=0$, it becomes the second circle $\therefore$ it is true for all values of $p$

Hence, the correct answer is option: 'A'.

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