Centre and radius of circle in x^2+y^2-8x-4y-5=0
Answers
Answered by
3
adding and subtracting 20 we get
x^2+y^2-8x-4y-5+20-20=0
(x-4)^2+(y-2)^2=5^2
hence centre is (4,2) radius=5
x^2+y^2-8x-4y-5+20-20=0
(x-4)^2+(y-2)^2=5^2
hence centre is (4,2) radius=5
Similar questions