A diameter of the circle $x^{2}+y^{2}=2x+6y-6$ is a chord of the another circle with centre $(2,1)$. The radius of this circle $C$ is $(in\ units) $

By completing the square for x^2 + y^2 = 2x + 6y - 6, the first circle has center (1, 3) and radius 2. The endpoints of the diameter of this first circle are (3, 3) and (-1, 3). This diameter acts as a chord for the second circle centered at (2, 1). The distance from the second circle's center (2, 1) to this chord (the line y = 3) is 2. Using the perpendicular distance from the center to the chord, the radius R of the second circle is found by R^2 = 2^2 + (half the chord length)^2. This gives R^2 = 4 + 5, so R equals 3.

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