Question

Find the centre and radius of the circle $x^2+y^2-8x+10y-12=0$

Answer 1

$x^2+y^2-8x+10y-12=0$

$\Rightarrow (x^2-8x)+(y^2+10y)=12$

$\Rightarrow \{x^2-2\left(x\right)\left(4\right)+4^2\}+\{y^2+2\left(y\right)\left(5\right)+5^2\}-16-25=12$

$\Rightarrow {(x-4)}^2+{(y+5)}^2=53$

$\Rightarrow {(x-4)}^2+{\{y-(-5)\}}^2={\left(\sqrt{53}\right)}^2$

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 $\sqrt{53}$