JEE Main · 2009 · 12 Apr 2009 · Offline · Circle

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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

(A)all values of $p$ ✓ Correct
(B)all except one value of $p$
(C)all except two values of p
(D)exactly one value of $p$

Answer: (A)

Solution

Every circle through the common points

P

and

Q

of the two given circles can be written as

S + λ S^{'} = 0

. Forcing this to pass through

(1, 1)

fixes the parameter through

S (1, 1) + λ S^{'} (1, 1) = 0

, that is

(7 + 2 p) + λ (6 - p^{2}) = 0

, from which a suitable

λ

can always be found. So whatever the value of

p

, the family contains a circle through

P

Q

and

(1, 1)

, and the property holds for all values of

p

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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