Question

Find the coordinates of the point equidistant from the axis and lying on the circle x square + y square - 6 x minus 2 y + 6 is equal to zero

Answer 1

${x}^{2} + {y}^{2} - 6x - 2y + 6 = 0 \\ ({x}^{2} - 6x) + ( {y}^{2} - 2y) + 6 = 0 \\ x(x - 6) + y(y - 2) + 6 = 0 \\ x + y +6 = - x + 6 - y + 2 \\ 2x + 2y = 2 \\ x + y = 1 \\ x = 1 - y \\ x + y = 1 \\ 1 - y + y = 1 \\ 1 = 1$

therefore the coordinates of the points are (1,1)

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