Find the equation of the circle with center ( | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Find the equation of the circle with center $(2, 1)$ and touching the $X$-axis.

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