one of the diameters of the circle x square+y square-12x+4y+6=0 is given by
a.x+y=0 b.x=y
c.x+3y=0 d.3x+2y=0
 

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{x}}^2+{\mathrm{y}}^2-12\mathrm{x}+4\mathrm{y}+6=0 \\ \mathrm{Centre} \mathrm{of} \mathrm{circle} \mathrm{is} \left(6,-2\right) \\ \mathrm{Now} \mathrm{dinameter} \mathrm{of} \mathrm{circle} \mathrm{passes} \mathrm{through} \mathrm{centre}, \mathrm{hence} \mathrm{centre} \mathrm{must} \mathrm{lie} \\ \mathrm{on} \mathrm{diameter}. \\ \mathrm{Of} \mathrm{given} \mathrm{options} \left(6,-2\right) \mathrm{lies} \mathrm{only} \mathrm{on} \mathrm{x}+3\mathrm{y}=0 \left(6-3\times 2=0\right) \\ \mathrm{Hence} \mathrm{option}\left(\mathrm{c}\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

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