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anonymous
 4 years ago
how to find the center and radius of x^2+y^2x+2y+1=0?
anonymous
 4 years ago
how to find the center and radius of x^2+y^2x+2y+1=0?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0centre at: \( (\frac 12,1) \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0radius is \(\frac 12 \) units.

xkehaulanix
 4 years ago
Best ResponseYou've already chosen the best response.1FoolForMath 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 nonsquared 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 equationboth 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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