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

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Solution

The correct option is B
Centre (1, 2) and since circle touches x-axis, therefore, radius is equal to 2.
Hence the equation is $(x - 1)^{2} + (y - 2)^{2} = 2^{2}$
$\Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
Trick: The only circle is $x^{2} + y^{2} - 2 x - 4 y + 1 = 0,$ whose centre is (1, 2).

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

The correct option is B Centre (1, 2) and since circle touches x-axis, therefore, radius is equal to 2. Hence the equation is (x−1)2+(y−2)2=22 ⇒x2+y2−2x−4y+1=0 Trick: The only circle is x2+y2−2x−4y+1=0, whose centre is (1, 2).

The correct option is B
Centre (1, 2) and since circle touches x-axis, therefore, radius is equal to 2.
Hence the equation is $(x - 1)^{2} + (y - 2)^{2} = 2^{2}$
$\Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
Trick: The only circle is $x^{2} + y^{2} - 2 x - 4 y + 1 = 0,$ whose centre is (1, 2).