question_answer

The equation of the pair of straight lines parallel to x-axis and touching the circle \[{{x}^{2}}+{{y}^{2}}-6x-4y-12=0\] [Kerala (Engg.) 2002]

A)            \[{{y}^{2}}-4y-21=0\]        

B)            \[{{y}^{2}}+4y-21=0\]

C)            \[{{y}^{2}}-4y+21=0\]       

D)            \[{{y}^{2}}+4y+21=0\]

Correct Answer: A

Solution :

           Let the lines are \[y={{m}_{1}}x+{{c}_{1}}\] and \[y={{m}_{2}}x+{{c}_{2}}\] since pair of straight lines parallel to x-axis,   \\[{{m}_{1}}={{m}_{2}}=0\]            and the lines will be \[y={{c}_{1}}\]and \[y={{c}_{2}}\]            Given circle is \[{{x}^{2}}+{{y}^{2}}-6x-4y-12=0\], centre (3, 2) and radius = 5.            Here, the perpendicular drawn from centre to the lines are CP and \[C{P}'\].            \[CP=\frac{2-{{c}_{1}}}{\sqrt{1}}=\pm 5\] Þ \[2-{{c}_{1}}=\pm 5\]            \[{{c}_{1}}=7\] and \[{{c}_{1}}=-3\]            Hence the lines are            \[y-7=0,\,y+3=0\] i.e.,\[(y-7)\,(y+3)=0\]            \ Pair of straight lines is \[{{y}^{2}}-4y-21=0\].