The radius of a circle whose center is (2, 1) | vedclass

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(D) The given circle is $x^2 + y^2 - 2x - 6y + 6 = 0$. Completing the square,we get $(x - 1)^2 + (y - 3)^2 = 4$. The center of this circle is $(1, 3)$

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(D) The given circle is $x^2 + y^2 - 2x - 6y + 6 = 0$. Completing the square,we get $(x - 1)^2 + (y - 3)^2 = 4$. The center of this circle is $(1, 3)$ and its radius is $2$. Since the chord of the required circle is a diameter of this circle,the chord lies on the line passing through the center $(1, 3)$ and its endpoints are the points where the chord intersects the circle. However,the problem states the chord is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$. Thus,the chord is the line segment connecting the endpoints of the diameter of the given circle. The endpoints of the diameter of $x^2 + y^2 - 2x - 6y + 6 = 0$ are $(1, 1)$ and $(1, 5)$. The required circle has center $C(2, 1)$ and passes through the points $(1, 1)$ and $(1, 5)$. The radius $r$ is the distance from $(2, 1)$ to $(1, 1)$: $r = \sqrt{(2 - 1)^2 + (1 - 1)^2} = \sqrt{1^2 + 0^2} = 1$. Alternatively,checking the distance to $(1, 5)$: $r = \sqrt{(2 - 1)^2 + (1 - 5)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{17}$. Re-evaluating the problem statement: If the chord is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$,the chord is the line segment between $(1, 1)$ and $(1, 5)$,which is the vertical line $x = 1$. The distance from the center $(2, 1)$ to the line $x = 1$ is $d = |2 - 1| = 1$. The radius $r$ of the circle is the distance from the center $(2, 1)$ to any point on the chord,e.g.,$(1, 1)$. $r = \sqrt{(2 - 1)^2 + (1 - 1)^2} = 1$.

The radius of a circle whose center is $(2, 1)$ and one of its chords is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is:

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