Question:
Find the centre and radius of the circle $x^{2}+y^{2}-8 x+10 y-12=0$
Solution:
The equation of the given circle is $x^{2}+y^{2}-8 x+10 y-12=0$.
$x^{2}+y^{2}-8 x+10 y-12=0$
$\Rightarrow\left(x^{2}-8 x\right)+\left(y^{2}+10 y\right)=12$
$\Rightarrow\left\{x^{2}-2(x)(4)+4^{2}\right\}+\left\{y^{2}+2(y)(5)+5^{2}\right\}-16-25=12$
$\Rightarrow(x-4)^{2}+(y+5)^{2}=53$
$\Rightarrow(x-4)^{2}+\{y-(-5)\}^{2}=(\sqrt{53})^{2}$, which is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $h=4, k=-5$, and $r=\sqrt{53}$.
Thus, the centre of the given circle is $(4,-5)$, while its radius is $\sqrt{53}$.