• anonymous
how to find the center and radius of x^2+y^2-x+2y+1=0?
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
centre at: \( (\frac 12,-1) \)
  • anonymous
radius is \(\frac 12 \) units.
  • xkehaulanix
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.

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