could someone help me please find the centre and radius of the circle x^2-10x+y^2+2y+25=0?

Answer 1

centeris
(5,-1)
radius
=(25+1-25)^1/2
=1

Answer 2

The equation of a circle with center (a,b) and radius r is
(x-a)^2 + (y-b)^2 = r^2

Do you know how to complete the square? The easiest way is to complete the squares for x and y, and move the constant terms to the right of =.

x^2-10x +y^2+2y +25=0

(x-5)^2 + (y+1)^2 + 25 = 25 + 1

(x-5)^2 + (y+1) ^2 = 1

So the center is (5,-1) and radius is 1

Answer 3

The given expression can be written as
[x-5]^2+[y+1]^2=1
This is the equation of a cicle with centre at[5,-1] and radius 1

Answer 4

The classic equation for a circle is: (x-a)^2+(y-b)^2=r^2
Now your problem: [x^2-2*5x]+[y^2-(-2y)]+25=0 or (x^2-2*5x+5*5)+(y^2-(-2y)+1*1)-1=0
Or (x-5)^2+(y+1)^2=1 (check me out, no errors?). therefore [5,-1] is center point, radius=1

Answer 5

compare with x^2+y^2+2gx+2fy+c+0
centre=(-g,-f)=(5,-1)
radius=rt(g^2+f^2-c)
=rt(25+1-25)=1

Answer 6

[x-5]^2+[y+1]^2=1
5,1