lf $(1, 2), (4, 3)$ are the limiting points of a coaxial system then the equation of the circle in its conjugate system having minimum area is

x^{2} + y^{2} - 2 x - 4 y + 5 = 0

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x^{2} + y^{2} - 8 x - 6 y + 25 = 0

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x^{2} + y^{2} - 5 x - 5 y + 10 = 0

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x^{2} + y^{2} + 5 x + 5 y - 10 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 5 x - 5 y + 10 = 0$
Circle of conjugate system with minimum radius will be the circle having
limiting points as diameter.
Thus required equation is given by, $(x - 1) (x - 4) + (y - 2) (y - 3) = 0$
$\Rightarrow x^{2} + y^{2} - 5 x - 5 y + 10 = 0$

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