Trigonometric EquationHard
Question
If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y + 2p - 5 = 0 and x2 + y2 + 2x + 2y - p2 = 0, then there is a circle passing through P, Q and (1, 1) for
Options
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 = x2 + y2 + 3x + 7y + 2p - 5 = 0
S′ = x2 + y2 + 2x + 2y - p2 = 0
Equation of required circle is S + λs′ = 0
As it passes through (1, 1) the value of
If 7 + 2p = 0, it becomes the second circle
∴ it is true for all values of p
S′ = x2 + y2 + 2x + 2y - p2 = 0
Equation of required circle is S + λs′ = 0
As it passes through (1, 1) the value of
If 7 + 2p = 0, it becomes the second circle
∴ it is true for all values of p
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