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Question

A circle touches the lines $y equals fraction numerator x over denominator square root of 3 end fraction comma y equals x square root of 3$ and has unit radius. If the centre of this circle lies in the first quadrant, then one possible equation of this circle is

$x^{2} + y^{2} - 2 x (\sqrt{3} + 1) - 2 y (\sqrt{3} + 1) + 8 + 4 \sqrt{3} = 0$
$x^{2} + y^{2} - 2 x (\sqrt{3} + 1) - 2 y (\sqrt{3} + 1) + 5 + 4 \sqrt{3} = 0$
$x^{2} + y^{2} - 2 x (\sqrt{3} + 1) - 2 y (\sqrt{3} + 1) + 7 + 4 \sqrt{3} = 0$
$x^{2} + y^{2} - 2 x (\sqrt{3} + 1) - 2 y (\sqrt{3} + 1) + 6 + 4 \sqrt{3} = 0$

The correct answer is: $x to the power of 2 end exponent plus y to the power of 2 end exponent minus 2 x left parenthesis square root of 3 plus 1 right parenthesis minus 2 y left parenthesis square root of 3 plus 1 right parenthesis plus 7 plus 4 square root of 3 equals 0$

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