Complete the square and write the equation in standard form. Then give the center and radius of the circle. x^2 + y^2 - 10x - 8y + 29 = 0 A. (x - 5)^2 +(y - 4)^2 = 12 (5, 4), r = 12 B. (x - 5)^2 +(y - 4)^2 = 12 (5, 4), r = 2 sq rt 3 C. (x + 5)^2 +(y + 4)^2 = 12 (-5, -4), r = 2 sq rt 3

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Complete the square and write the equation in standard form. Then give the center and radius of the circle. x^2 + y^2 - 10x - 8y + 29 = 0 A. (x - 5)^2 +(y - 4)^2 = 12 (5, 4), r = 12 B. (x - 5)^2 +(y - 4)^2 = 12 (5, 4), r = 2 sq rt 3 C. (x + 5)^2 +(y + 4)^2 = 12 (-5, -4), r = 2 sq rt 3

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is this calculis
Yes pre-calc
start with \[x^2 + y^2 - 10x - 8y + 29 = 0\] then group the terms together as \[x^2-10x+y^2-8y=-29\] to complete the square, take half the coefficient of the \(x\) and \(y\) term and write \[(x-5)^2+(y-8)^2=-29+5^2+4^2\] then compute the number on the right

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ooh thats y im in geometry srry
you get \[(x-5)^2+(y-4)^2=12\] so the circle has center at \((5,4)\) and radius \(\sqrt{12}\) or \(2\sqrt{3}\) if you prefer
thank you
yw
see u marrow alyssa :) bye

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