If P and Q are the points of intersection of the circles x²+y²+3 x+7 y+2 p-5=0 and x²+y²+2 x+2 y-p²=0 then there is a circle passing through P, Q and (1,1) for: [AIEEE 2009]

Correct Answer is : all except one value of ppp

Correct Answer is : all except one value of ppp

Test Index

Circles Part 1

Section: Mathematics

Question : 81 of 100

Marks: +1, -0

P

Q

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

x^2+y^2+2 x+2 y-p^2=0

P, Q

(1,1)

Solution:

The given circles are

S_1 \equiv x^2+y^2+3 x+7 y+2 p-5=0

.....(1)

S_2 \equiv x^2+y^2+2 x+2 y-p^2=0

.....(2)

\therefore

Equation of common chord

P Q

S_1-S_2=0

\Rightarrow L \equiv x+5 y+p^2+2 p-5=0

\Rightarrow

Equation of circle passing through

P

and

Q

S_1+\lambda L=0

\Rightarrow (x^2+y^2+3 x+7 y+2 p-5) +\lambda

(x+5 y+p^2+2 p-5) =0

As it passes through

(1,1)

, therefore

\;\Rightarrow (7+2 p)+\lambda (2 p+p^2+1) =0

\;\Rightarrow \lambda = -\; \frac{2 p+7}{p+1}

which does not exist for

p=-1

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