Question

If one end of a diameter of a circle is $x^2+y^2+8x+2y+12=0$ is $(1,2)$ then find the co-ordinates of the other end of the diameter.

Answer 1

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

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

Its centre is $C(-g,-f)=C(-4,-1)$

Let $A(1,2)$ and $B(\alpha ,\beta )$ be the ends of the diameter, then $C(-4,-1)$ is the midpoint of the segment $AB$

$\therefore \frac{1+\alpha }{2}=-4$ and $\frac{2+\beta }{2}=-1$

$\Rightarrow 1+\alpha =-8$ and $2+\beta =-2$

$\Rightarrow \alpha =-8-1=-9$ and $\beta =-2-2=-4$

Hence, the co-ordinates of the other end of the diameter are $(-9,-4)$