Find equation at line passing through mid-point of intercept by circle $x^{2} + y^{2} - 4 x - 6 y = 0$ on co-ordinate axis.

3 x + 2 y - 6 = 0

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3 x + y - 6 = 0

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3 x + 4 y - 12 = 0

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3 x + 2 y - 12 = 0

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Solution

The correct option is A $3 x + 2 y - 6 = 0$
$\begin{matrix} x^{2} + y^{2} - 4 x - 6 y = 0 {(x - 2)}^{2} + {(y - 3)}^{2} = 13 c e n t r e (2, 3) R a d i u s = \sqrt{13} i n t e r sec t i o n b e t w e e n t h e c i r c l e a n d x - a x i s . {(x - 2)}^{2} + {(0 - 3)}^{2} = 13 {(x - 2)}^{2} = 4 x - 2 = \pm 2 x = 4, 0 A (0, 0), B (4, 0) I n t e r sec t i o n b e t w e e n y - a x i s a n d c i r c l e . {(0 - 2)}^{2} + {(y - 3)}^{2} = 13 y - 3 = \pm 3 y = 0, 6 c (0, 0), D (0, 6) S o, e q u a t i o n o f l i n e j o i n i n g t h e m i d p o i n t s o f t h e x - i n t e r c e p t a n d y - i n t e r c e p t a r e y - 0 = \frac{3 - 0}{0 - 2} (x - 2) - 2 y = 3 x - 6 3 x + 2 y - 6 = 0 3 x + 2 y - 6 = 0 \end{matrix}$

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