JEE AdvancedCircleMedium
Question
Two circles x2 + y2 = 6 and x2 + y2 - 6x + 8 = 0 Then the equation of the circle through their points of intersection and point (1, 1) is
Options
A.x2 + y2 - 6x + 4 = 0
B.x2 + y2 - 3x + 1 = 0
C.x2 + y2 - 4y + 2 = 0
D.None of these
Solution
The required equation of circle is, S1 + λ(S2 - S1) = 0
∴ (x2 + y2 - 6) + λ (-6x + 14) = 0
Also passing throught (1, 1)
⇒ - 4 + λ (8) = 0
⇒ λ =
∴ Required equation of circle is,
x2 + y2 - 6 - 3x + 7 = 0
or x2 + y2 - 3x +1 = 0
∴ (x2 + y2 - 6) + λ (-6x + 14) = 0
Also passing throught (1, 1)
⇒ - 4 + λ (8) = 0
⇒ λ =
∴ Required equation of circle is,
x2 + y2 - 6 - 3x + 7 = 0
or x2 + y2 - 3x +1 = 0
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