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

  • Option 1)

    all except one value of  p

  • Option 2)

    all except two values of  p

  • Option 3)

    exactly one value of  p

  • Option 4)

    all values of p

 

Answers (1)

As we learnt in

Family of circle -

S+KL= 0

- wherein

Equation of the family of circles passing through point of intersection S=0 \, and\, line\, L=0.

 

                                                            x^{2}+y^{2}+3x+7y+2p-5 = 0

                                                    {x^{2}+y^{2}+2x+2y-p^{2}=0}\\ (-)

                                                            \overline {{x+5y+\left ( p^{2}+2p-5 \right ) = 0}}

Equation of common chord family of circles is

\left ( x^{2}+y^{2}+3x+7y+2p-5 \right )+ \lambda \left ( x+5y+p^{2}+2p-5 \right )= 0

It passes through (1,1)

\left ( 7+2p \right )+ \lambda \left ( 2p+p^{2}+1 \right )= 0

\lambda = \frac{-\left ( 2p+7 \right )}{\left ( p+1 \right )^{2}}

So, p\neq 1\:\:\: only\:case.


Option 1)

all except one value of  p

This option is correct.

Option 2)

all except two values of  p

This option is incorrect.

Option 3)

exactly one value of  p

This option is incorrect.

Option 4)

all values of p

This option is incorrect.

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