If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\), then the other end of the…
If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\), then the other end of the diameter is
\((0,1)\)
\((1,1)\)
\((1,2)\)
\((1,0)\)
Solution
Given, circle \(x^2+y^2-4 x-6 y+11=0\)
\(\Rightarrow \quad\) Centre \(=(2,3)\)
One end diameter \(=(3,4)\)
Let other end be \((h, k)\)
So, \(\frac{h+3}{2}=2, \frac{k+4}{2}=3\)
\(\Rightarrow h=1, k=2\)