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The equation of the circle which has normal $(x - 1) \times (y - 2) = 0$ and a tangent $3 x + 4 y = 6$ is

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

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

$x^{2} + y^{2} = 5$

${(x - 3)}^{2} + {(y - 4)}^{2} = 5$

answer is A.

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Detailed Solution

x-1=0,y-2=0

centre is(1,2),radius is perpendicular distance from centre to tangent3x+4y=6

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