Question

If one end of a diameter of a circle is $x^2+y^2-4x-6y+11=0$ is $(3,4)$ then find the co-ordinates of the other end of the diameter.

Answer 1

The given circle is $x^2+y^2-4x-6y+11=0$ .......$\left(1\right)$

is of the form $x^2+y^2+2gx+2fy+c=0$

Its centre is $C(-g,-f)=C(2,3)$

Let $A(3,4)$ and $B(\alpha ,\beta )$ be the ends of the diameter, then $C(2,3)$ is the midpoint of the segment $AB$

$\therefore \frac{3+\alpha }{2}=2$ and $\frac{4+\beta }{2}=3$

$\Rightarrow 3+\alpha =4$ and $4+\beta =6$

$\Rightarrow \alpha =4-3=1$ and $\beta =6-4=2$

Hence, the coordinates of the other end of the diameter are $(1,2)$