Question

If one end of a diameter of a circle is $x^2+y^2-8x-12y+22=0$ is $(2,3)$ then find the co-ordinates of the other end of the diameter.

Answer 1

The given circle is $x^2+y^2-8x-12y+22=0$ .......$\left(1\right)$

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

Its centre is $C(-g,-f)=C(4,6)$

Let $A(2,3)$ and $B(\alpha ,\beta )$ be the ends of the diameter, then $C(4,6)$ is the midpoint of the segment $AB$

$\therefore \frac{2+\alpha }{2}=4$ and $\frac{3+\beta }{2}=6$

$\Rightarrow 2+\alpha =8$ and $3+\beta =12$

$\Rightarrow \alpha =8-2=6$ and $\beta =12-3=9$

Hence, the coordinates of the other end of the diameter are $(6,9)$