IIT-JEE 1980 | Circle Question 93 | Mathematics

IIT-JEE 1980

MCQ (Single Correct Answer)

-0.25

Two circles $${x^2} + {y^2} = 6$$ and $${x^2} + {y^2} - 6x + 8 = 0$$ are given. Then the equation of the circle through their points of intersection and the point (1, 1) is

$${x^2} + {y^2} - 6x + 4 = 0$$

$${x^2} + {y^2} - 3x + 1 = 0$$

$${x^2} + {y^2} - 4y + 2 = 0$$

none of these

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