The equation of the circle having the lines $y^{2} - 2 y + 4 x - 2 x y = 0$ as its normals and passing through the point $(2, 1)$ is

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

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

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x^{2} + y^{2} + 2 x + 4 y - 13 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$y^{2} - 2 y + 4 x - 2 x y = 0$

$y (y - 2) - 2 x (y - 2) = 0$

$y = 2 x, y = 2$ are normal to the circle and every normal passes through the center

Intersection of normal will give us center

$y = 2 x, y = 2 ⟹ x = 1$

C = (1,2) and P = (2,1)

CP must be equal to radius

$⟹ r = \sqrt{(2 - 1)^{2} + (1 - 2)^{2}} = \sqrt{2}$

Equation of circle having center (a,b) and radius r is $(x - a)^{2} + (y - b)^{2} = r^{2}$

$⟹ (x - 1)^{2} + (y - 2)^{2} = 2$

$⟹ x^{2} + y^{2} - 2 x - 4 y + 3 = 0$

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The equation of the circle having the lines y2−2y+4x−2xy=0 as its normals and passing through the point (2,1) is

The equation of the circle having the lines $y^{2} - 2 y + 4 x - 2 x y = 0$ as its normals and passing through the point $(2, 1)$ is