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well the general form of the circle you need is \[(x - h)^2 + (y - k)^2 = r^2\] so you need to complete the square in both x and y... can you do that..?
what does k stand for and how do I find the height in the eqaution?
(h, k) is the centre of the circle and r is the radius
okay so I need to find x first?
no you have \[x^2 - 4x \] you need to add a value so it forms a perfect square e.g.\[(x - a)^2 = x^2 -2ax + a^2 \] this is called completing the square.. ay thoughts...
so you need to compare 2ax and 4x any thoughts on the value of a...?
im guessing 2
great so 2^2 = 4 so add 4 to both sides of the equation to keep it in balance. now the equation can be written as \[(x - 2)^2 +y^2 + 8y = 0\] next complete the square in y... \[(y + a)^2 =y^2 + 2ay + a^2 \] so compare 2ay with 8y what is the value of a..?
yes... so 4^2 = 16 again add 16 to both sides of the equation... and you get \[(x -2)^2 + (y + 4)^2 = 16\] now you can identify the radius and centre
is it D?
the expanded form of the equation, just to check is \[x^2 - 4x + 4 + y^2 _ 8y + 16 = 16 ~~or ~~x62 -4x + y^2 + 8y + 20 = 16\] subtract 20 from both sides of the equation and you get \[x^2 -4x + y^2 + 8y = -4\] the original equation
no it's not... remember the equation is \[(x - h)^2 + (y - k)^2 = r^2\] (h, k) is the centre r is the radius... so take the square root to find r
the centre is correct..
that's correct now you have all the information you need
so B is the correct answer correct?
i would fsn u but i already am