Explanation:
A Given equation of circle is,
\(x^2+y^2-6 x-4 y+11=0\)
Hence, the centre of circle is \(\mathrm{O}(3,2)\)
The co-ordinate point \(\mathrm{P}(4,3)\) is an end diameter which lies on the circle.
\(\mathrm{O}_{(3,2)}^{\mathrm{Q}(\mathrm{x}, \mathrm{y})}\)
Let, \(\mathrm{Q}(\mathrm{x}, \mathrm{y})\) the other end of diameter So, centre \(\mathrm{O}(3,2)\) is the mid point of \(\mathrm{PQ}\). \(\therefore 3=\frac{4+\mathrm{x}}{2}\) and \(2=\frac{3+\mathrm{y}}{2}\)
\(\mathrm{x}=2 \quad \text { and } \mathrm{y}=1\)So, \(\mathrm{Q}(2,1)\) is the other end of diameter.