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(B) The centre of the circle is $(h, k) = (1, 2)$.Since the circle touches the $x$-axis,the radius $r$ is equal to the absolute value of the $y$-coord

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The centre of the circle is $(h, k) = (1, 2)$.Since the circle touches the $x$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre,so $r = |2| = 2$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 1)^2 + (y - 2)^2 = 2^2$.Expanding this,we get $(x^2 - 2x + 1) + (y^2 - 4y + 4) = 4$.Simplifying,we get $x^2 + y^2 - 2x - 4y + 1 = 0$.

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

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