Find the centre and radius of the circle x2+y2−8x+10y−12=0
x2+y2−8x+10y−12=0
⇒(x2−8x)+(y2+10y)=12
⇒{x2−2(x)(4)+42}+{y2+2(y)(5)+52}−16−25=12
⇒(x−4)2+(y+5)2=53
⇒(x−4)2+{y−(−5)}2=(√53)2
which is of the form (x−h)2+(y−k)2=r2
where h=4,k=−5 and r=√53
Thus the centre of the given circle is (4,−5) while its radius is √53