SOLUTION: i am trying to find the center and radius of X2+Y2+4X-8Y+13=0 So far i think its (x+2)^2+(y-4)^2=-5^2 meaning the center would be (-2,4) and radius -5

Question 643602: i am trying to find the center and radius of X2+Y2+4X-8Y+13=0
So far i think its (x+2)^2+(y-4)^2=-5^2 meaning the center would be (-2,4) and radius -5
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
You have the center correct but not the radius.

We start by completing the squares for both the x terms and the y terms. Usually this tarts with "moving" the constant term to the other side. Subtracting 13 from each side we get:

Then we rearrange the terms so that the x terms are together and the y terms are together:

To complete the square for the x terms we take half the coefficient of x, 4, and square it. Half of 4 is 2 and 2 squared is, coincidentally, 4. We do the same for the y term. Half of -8 is -4 and -4 squared is 16. So we will add a 4 and a 16 to each side:

Next we simplify the right side and rewrite the left side as binomial squares:

Last of all we rewrite the right side as a perfect square:

So the center is (-2, 4) but the radius is .
RELATED QUESTIONS
x^2+y^2-4x+6y+4=0 i need to find the center, and... (answered by tutor_paul)

given the equation of a circle x^2+y^2-10x+4y+13=0 find its center and its radius. im... (answered by dabanfield)

I need to find the x-intercept and the y- intercept, radius, center and graph the circle. (answered by ewatrrr)

Please let me know if this is correct. Question: Given the equation of a circle is... (answered by RAY100)

given the equation of a circle x^2+y^2-10x+4y+13=0. find its center and radius. i am... (answered by Theo)

Find the center and radius of the circle given by... (answered by stanbon)

Find the center and radius of the circle: x^2 + y^2 - 6x + 8y + 21 =... (answered by Nate)

Find the center and radius of the circle... (answered by Fombitz,solver91311)

1 Use the discriminant to determine whether the quadratic equation has two unequal real... (answered by user_dude2008,stanbon)