SOLUTION: I tried to solve this problem but I didnt come to far, could someone help me with it? Problem; Given that the equation of a circle is x^2+y^2-10x+4y+13=0,find �ts center and its

Question 206830: I tried to solve this problem but I didnt come to far, could someone help me with it?
Problem; Given that the equation of a circle is x^2+y^2-10x+4y+13=0,find �ts center and its radius.
The way I tried to solve it;
(x-h)^2+(y-k)^2=r^2
(x^2-10x+9)+(y^2+4y+1)= -13+9+1
(I think I did something totally wrong here)
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
you need to COMPLETE THE SQUARE for the x's and y's

you're on the right track, but you need different constant terms

the first order terms (-10x and 4y) determine the constants

x^2 - 10x + 25 = (x - 5)^2

y^2 +4y + 4 = (y + 2)^2

(x^2 - 10x + 25) + (y^2 + 4y + 4) = -13 + 25 + 4 = 16

(x - 5)^2 + (y + 2)^2 = 4^2
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