Consider a family of circle passing through the point of intersection of lines $\sqrt{3} (y - 1) = x - 1 and y - 1 = \sqrt{3} (x - 1)$ and having its centre on the acute angle bisector of the given lines. Then the common chords of each member of the family and the circle $x^{2} + y^{2} + 4 x - 6 y + 5 = 0$ are concurrent at the point

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$(\frac{3}{2}, \frac{3}{2})$

$(\frac{1}{2}, \frac{1}{2})$

$(- \frac{1}{2}, - \frac{1}{2})$

$(\frac{1}{2}, \frac{3}{2})$

answer is B.

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Detailed Solution

Point of intersection of lines (1, 1). Slopes of lines are $30^{\circ} and 60^{\circ}$ so acute angle bisector having $θ = 45^{\circ} \Rightarrow y = x$ equation of circle
$\begin{array}{l} (x - a)^{2} + (y - a)^{2} = (1 - a)^{2} + (1 - a)^{2} \\ \Rightarrow x^{2} + y^{2} - 2 ax - 2 ay + 4 a - 2 = 0 \end{array}$
Common chord with given circles is
$\begin{array}{l} x^{2} + y^{2} + 4 x - 6 y + 5 = 0 \\ (4 + 2 a) x + y (2 a - 6) + (7 - 4 a) = 0 \\ \Rightarrow (4 x - 6 y + 7) + 2 a (x + y - 2) = 0 \\ \Rightarrow P.O.I (\frac{1}{2}, \frac{3}{2}) \end{array}$