If one of the diameters of the circle \(x^{2}+y^{2}-2 x-6 y+6=0\) is a chord to the circle with centre \((2,1)\), then the radius of the circle is \(\ldots\) (a) 3 (b) \(\sqrt{3}\) (c) 2 (d) \(\sqrt{2}\)

Rewrite the given equation of the circle in the standard form \((x-a)^2 + (y-b)^2 = r^2\). We have: \(x^2 -2x + y^2 -6y + 6 = 0\) Add 1 and 9 on both sides to complete the squares for x and y variables respectively, \((x^2 -2x+1) + (y^2 -6y +9) = 4\) \((x-1)^2 + (y-3)^2 = 4\) Now we can see that the center of the first circle is \((1,3)\), and its radius is \(r = 2\).

By definition, a diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circle. Let's assume that the x-axis is the diameter we are working with. Since the center of the first circle is on the x-axis, all we need to do is find the y-coordinate of its endpoints. The endpoints of the diameter have the same x-coordinate as the center, \(x=1\). To find the y-coordinates, we can plug these values into the equation of the first circle, \((1-1)^2 + (y-3)^2 = 2^2\) \((y-3)^2 = 4\) Thus, \(y = 1\) or \(y = 5\). So the endpoints of the diameter are \((1,1)\) and \((1,5)\).

The distance between the endpoints of the diameter is the length of this chord of the second circle. Using the distance formula, the length (\(L\)) of the chord is: \(L = \sqrt{(1-1)^2 + (5-1)^2} = \sqrt{16} = 4\)

Let the second circle have the center \((2,1)\) and a radius \(r_2\). We construct a perpendicular bisector of the chord (diameter of the first circle), which passes through the center of the second circle and is of length \(L/2\). Let this point be \((x_m, y_m)\). Since the midpoint of the chord has coordinates \(((1+1)/2,(1+5)/2)=(1,3)\), and we know that this bisector passes through \((2,1)\) and has a length \(L/2 = 4/2 = 2\), we can determine the length between the center of the second circle and this perpendicular bisector using the Pythagorean theorem: \(r_2^2 = 2^2 + (\text{distance from }(2,1) \text{ to }(1,3))^2\) \(r_2^2 = 4 + (1^2 + 2^2)\) \(r_2^2 = 4 + 5\) \(r_2^2 = 9\) \(r_2 = \sqrt{9} = 3\) Thus, the radius of the second circle is 3. The correct answer is (a) 3.