The equation of the circle which touches x-axis and whose center is (1,2) is

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

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x^{2} + y^{2} - 2 x - 4 y + 1 = 0

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x^{2} + y^{2} + 2 x + 4 y + 1 = 0

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

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Solution

The correct option is B $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
$∴ O \equiv (1, 2) r = 2 c m$
$h, k$
$∴$ Eqn of circle:
$(x - h)^{2} + (y - k)^{2} = r^{2}$
$(x - 1)^{2} + (y - 2)^{2} = 4$
$x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 4$
$x^{2} + y^{2} - 2 x - 4 y + 1 = 0$

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The equation of the circle which touches x-axis and whose center is (1,2) is

The correct option is B x2+y2−2x−4y+1=0∴O≡(1,2) r=2cm h,k∴ Eqn of circle:(x−h)2+(y−k)2=r2(x−1)2+(y−2)2=4x2+1−2x+y2+4−4y=4x2+y2−2x−4y+1=0

The correct option is B $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
$∴ O \equiv (1, 2) r = 2 c m$
$h, k$
$∴$ Eqn of circle:
$(x - h)^{2} + (y - k)^{2} = r^{2}$
$(x - 1)^{2} + (y - 2)^{2} = 4$
$x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 4$
$x^{2} + y^{2} - 2 x - 4 y + 1 = 0$