Question

If one end of a diameter of a circle is $x^2+y^2-6x-8y+12=0$ is $(4,6)$ then find the co-ordinates of the other end of the diameter.

Answer 1

The given circle is $x^2+y^2-6x-8y+12=0$ .......$\left(1\right)$

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

Its centre is $C(-g,-f)=C(3,4)$

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

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

$\Rightarrow 4+\alpha =6$ and $6+\beta =8$

$\Rightarrow \alpha =6-4=2$ and $\beta =8-6=2$

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