Given equation of the circle,
\[\begin{array}{*{35}{l}}
{{x}^{2}}~-\text{ }4x\text{ }+\text{ }{{y}^{2}}~\text{ }-6y\text{ }+\text{ }11\text{ }=\text{ }0 \\
{{x}^{2}}~\text{ }-4x\text{ }+\text{ }4\text{ }+\text{ }{{y}^{2}}~\text{ }-6y\text{ }+\text{ }9\text{ }+11\text{ }-\text{ }13\text{ }=\text{ }0 \\
\end{array}\]
the above equation can be written as
\[{{x}^{2}}~\text{ }-2\text{ }\left(2 \right)\text{ }x\text{ }+\text{ }{{2}^{2}}~+\text{ }{{y}^{2}}~\text{ }-2\text{ }\left( 3 \right)\text{ }y\text{ }+\text{ }{{3}^{2}}~+11\text{ }-\text{ }13\text{ }=\text{ }0\]
on simplifying we get
(x – 2)2 + (y – 3)2 = 2
(x – 2)2 + (y – 3)2 = (√2)2
Since, the equation of a circle having centre (h, k), having radius as r units, is
(x – h)2 + (y – k)2 = r2
We have centre = (2, 3)
The centre point is the mid-point of the two ends of the diameter of a circle.
Let the points be (p, q)
(p + 3)/2 = 2 and (q + 4)/2 = 3
p + 3 = 4 & q + 4 = 6
p = 1 & q = 2
Hence, the other ends of the diameter are (1, 2).