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If $L_{1}$ represents the radical axis of circles $x^{2} + y^{2} - 4 x - 6 y + 1 = 0 a n d$ $x^{2} + y^{2} - 2 x - 4 y - 1 = 0,$ and $L_{2}$ represents the radical axis of $x^{2} + y^{2} + 2 x + 2 y - 7 = 0 a n d x^{2} + y^{2} + x + y + 9 = 0$ , then ___

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$L_{1} i s p a r a l l e l t o L_{2}$

$L_{1} i s p e r p e n d i c u l a r t o L_{2}$

$L_{1} a n d L_{2} i n t e r s e c t a t a n a n g l e 30^{o}$

$L_{1} a n d L_{2} i n t e r s e c t a t (1, 7)$

answer is A.

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Detailed Solution

$Radical axis is S - S^{1} = 0 L_{1} \equiv (x^{2} + y^{2} - 4 x - 6 y + 1) - (x^{2} + y^{2} - 2 x - 4 y - 1) = 0 a n d L_{2} \equiv (x^{2} + y^{2} + 2 x + 2 y - 7) - (x^{2} + y^{2} + x + y + 9) = 0$

$i . e L_{1} \equiv x + y - 1 = 0 - (1) L_{2} \equiv x + y - 16 = 0 - (2) H e r e L_{1} is parallel to L_{2}$

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