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Find the centre of the circle with equation \(12x-12y-2x^2-2y^2=24\).

in Mathematics by (142 points) | 21 views

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The hint here is to first write the equation in the general form \(x^2+ y^2+ Ax +By +C = 0\) so that it can be easily written in the standard form \((x-a)^2 + (y-b)^2 = r^2\), here \((a,b)\) is the center of the circle and \(r\) is the radius.

\(12x -12y - 2x^2 -2y^2 = 24\\ -2x^2 -2y^2 +12x-12y -24 = 0 \\ x^2 + y^2 -6x +6y+12 =0 \\ x^2 -6x +y^2 +6y +12 = 0 \\ (x-3)^2 +(y+3)^2 -6 = 0 \\ (x-3)^2 + (y+3)^2 = 6\)

from the last line, we see that the center is \((3,-3)\) with radius \( \sqrt{6}\)

by Wooden (882 points)

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