Find the center and the radius of the circle with the equation: x^2-2x+y^2+4y+1=0?

Answer:Center (h,k) is (1,-2) and radius r = 2Step-by-step explanation:We need to find the center and radius of the circle of the given equation:[tex]x^2-2x+y^2+4y+1=0[/tex]We need to transform the above equation into standard form of circle[tex](x-h)^2 + (y-k)^2 = r^2[/tex]where(h,k) is the center of circle and r is radius of circle.Solving the given equation:[tex]x^2-2x+y^2+4y+1=0[/tex]Moving 1 to right side[tex]x^2-2x+y^2+4y=-1[/tex]Now making perfect square of x^2-2x and y^2+4yAdding +1 and +4 on both sides of the equation[tex]x^2-2x+1+y^2+4y+4=-1+1+4[/tex]Now, x^2-2x+1 is equal to (x-1)^2 and y^2+4y+2 =(y+2)^2[tex](x-1)^2+(y+2)^2=4[/tex]Comparing with standard equation of circle:[tex](x-h)^2 + (y-k)^2 = r^2[/tex]h = 1 , k =-2 and r =2 because r^2 =4 then r=2So, center (h,k) is (1,-2) and radius r = 2