Question

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

Answer 1

The given equation of circle is
$x^2+y^2-8x+10y-12=0\therefore (x^2-8x)+(y^2+10y)=12\Rightarrow [x^2-8x+(4{)}^2]+[y^2+10y+\left(5{)}^2\right]=12+(4{)}^2+(5{)}^2\Rightarrow (x-4{)}^2+(y+5{)}^2=12+16+25\Rightarrow (x-4{)}^2+(y+5{)}^2=53\Rightarrow (x-4{)}^2+(y+5{)}^2=(\sqrt{53}{)}^2$
comparing it with $(x-h{)}^2+(y-k{)}^2=r^2,$ we have
h = 4, k = - 5 and $r=\sqrt{53}$
Thus co-ordinates of the centre is (4, -5) and radius is $\sqrt{53}$