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$Let P, Q be the feet of perpendiculars drawn from point (1, 1) upon the lines$ $L_{1} \equiv x + 4 y - 12 = 0; L_{2} \equiv x - 4 y + 4 = 0 respectively. Let L_{3} = 0 and L_{4} = 0 be the$ $equations of the angle bisectors of L_{1} and L_{2} such that L_{3} is the line closer to$ $origin. Let R & S be the feet of perpendiculars drawn from (1, 1) upon the lines$ $L_{3} & L_{4} respectively. Let S_{1} = 0 be the equation of the circle passing through Q,$
$R & S and S_{2} = 0 be the equation of the circle passing through P, R & S . Then$
which of the following is/are TRUE?

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$S_{1} = 0 is given by x^{2} + y^{2} - 5 x - 3 y + 6 = 0$

$S_{2} = 0 is given by x^{2} + y^{2} + 5 x - 3 y - 6 = 0$

$Length of intercept on x -axis by S_{1} = \frac{1}{2}$

$Length of intercept on x = 0 by S_{2} = 1$

answer is A.

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Detailed Solution

$\begin{array}{l} x + 4 y - 12 = 0 \\ x - 4 y + 4 = 0 \end{array}$

Clearly A, B, P, Q, R, S are concyclic, with A and B as diametric and – points

$\begin{array}{l} Circle: (x - 1) (x - 4) + (y - 1) (y - 2) = 0 \\ x^{2} + y^{2} - 5 x - 3 y + 6 = 0 \\ Length of intercept on x -axis = 1 \end{array}$

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