SOLUTION: GIVE THE CENTER OF THE CIRCLE WITH EQUATION x^2+2x+y^2-10y+22=0.

Question 64976: GIVE THE CENTER OF THE CIRCLE WITH EQUATION x^2+2x+y^2-10y+22=0.
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
GIVE THE CENTER OF THE CIRCLE WITH EQUATION x^2+2x+y^2-10y+22=0.
The standard form of the equation of a circle is , (h,k)=center, r=radius
We complete the squares in order to put the equation in standard form.
x^2+2x+y^2-10y+22-22=0-22
x^2+2x+____+y^2-10y+_____=-22+_____+______
To complete the squares take (1/2) the coefficient of the middle term and add it to both sides of the equation.
x^2+2x+(2/2)^2+y^2-10y+(-10/2)^2=-22+(2/2)^2+(-10/2)^2
x^2+2x+(1)^2+y^2-10y+(-5)^2=-22+(1)^2+(-5)^2
x^2+2x+1+y^2-10y+25=-22+1+25
(x+1)^2+(y-5)^2=4
The center (h,k)=(-1,5)
If you were asked for the radius, it would be
Happy Calculating!!!
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