Question

If one end of a diameter of a circle is $x^2+y^2-6x-2y+4=0$ is $(-1,-2)$ then find the co-ordinates of the other end of the diameter.

Answer 1

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

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

Its centre is $C(-g,-f)=C(3,1)$

Let $A(-1,-2)$ and $B(\alpha ,\beta )$ be the ends of the diameter, then $C(3,1)$ is the midpoint of the segment $AB$

$\therefore \frac{-1+\alpha }{2}=3$ and $\frac{-2+\beta }{2}=1$

$\Rightarrow -1+\alpha =6$ and $-2+\beta =2$

$\Rightarrow \alpha =6+1=7$ and $\beta =2+2=4$

Hence, the co-ordinates of the other end of the diameter are $(7,4)$