Equation of circle passing through points of intersection of circles {x^2} + {y^2} - 6x + 8 = 0 and

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10-1.Circle and System of Circles

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The equation of circle passing through the points of intersection of circles ${x^2} + {y^2} - 6x + 8 = 0$ and ${x^2} + {y^2} = 6$ and point $(1, 1)$, is

A

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

B

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

C

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

D

None of these

(IIT-1980)

Solution

(b) Required equation is ${x^2} + {y^2} – 6x + 8 + \lambda ({x^2} + {y^2} – 6) = 0$. Now proceed.

Standard 11

Mathematics

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(JEE MAIN-2023)