Question

how to find the center and radius of x^2+y^2-x+2y+1=0?

Answer 1

centre at: \( (\frac 12,-1) \)

Answer 2

radius is \(\frac 12 \) units.

Answer 3

FoolForMath is right (I think).
You have to get the problem into a form you can recognize. The equation for a circle is
\[(x - a)^2 + (y - b)^2 = r^2\] Start by separating your equation by variables and the constant (1):
\[x^2 - x + y^2 + 2y = -1\] To factor the x and y terms, you need to finish the squares. Simplest way to do so is divide the non-squared x and y terms' coefficients (-1 and 2) by 2 and square them.
(-1/2)^2 = 1/4
(2/2)^2 = 1
Then add these into the equation-both sides of it to keep it balanced:\[x^2 - x + 1/4 + y^2 - 2y + 1 = -1 + 1/4 + 1\]Then simplify.\[x^2 - x + 1/4 + y^2 + 2y + 1 = 1/4\]Complete the squares on the left side of the equation (factor them):
\[(x - 1/2)^2 + (y + 1)^2 = 1/4\]The radius is of the form (a, b), which in this case is (1/2, -1). The radius is the square root of the right side of the equation, which in this case is 1/2.